
OH 





IBERT W. SMITH 

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library of congress, 



UNITED STATES OF AMERICA. 



ELEMENTARY 



MACHINE DESIGN 





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ALBERT W. SMITH 

Professor of Mechanical Engineering 
Leland Stanford Jr. University 



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CALIFORNIA 

STANFORD UNIVERSITY PRESS 

1895 






Copyrighted, i8 94 , by a. w. Smith. 






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* 



Professor John E. Sweet 



THIS BOOK IS DEDICATED 



AS AN EXPRESSION OF AFFECTION, AND IN 



ACKNOWLEDGMENT OF YEARS OF HELPFULNESS. 



PREFACE. 



One can never become a machine designer by studying a book. 
The true designer is one whose judgment is ripened by experience 
in constructing and operating machines. Mr. William B. Bement, 
a designer in the true sense, once said that he thought it useful to 
figure the strength of machine parts, because the results were sug- 
gestive to the designer. One may know thoroughly the laws which 
govern the transmission of energy; may understand much concern- 
ing the nature of constructive materials ; may know how to obtain 
results by mathematical processes ; and yet be unable to design a 
good machine. One needs also to know the thousand and one things 
connected with practice, which constantly modify design, so that one 
can take the results of computation and accept, reject, and modify, 
until the machine will, when constructed, do its required work 
satisfactorily and enduringly. The writer once heard Professor 
John E. Sweet say : " It is comparatively easy to design a good 
new machine, but it is very hard to design a machine which will be 
good when it is old." This quality of foresight only comes with 
long experience. 

There is, however, a certain part of the designer's mental equip- 
ment which may be furnished in the class-room, or by books. 
This is the writer's excuse for the following pages. 

Even Elementary Machine Design cannot be treated exhaust- 
ively. The kinds of machines are too numerous, and their differ- 
ences are too great. An effort is made here to suggest methods of 
reasoning, rather than to give rules. A knowledge of the usual 
university course in pure and applied mathematics is presupposed. 



VI PREFACE. 

The part upon " Motion in Machines " could not have been 
written without the use of the excellent book " The Mechanics of 
Machinery," by Prof. A. B. W. Kennedy. To him and to Prof. L. 
M. Hoskins, to whom the writer has so often gone for the help which 
never failed, grateful acknowledgment is here made. In several 
places acknowledgment is made to others ; yet the writer feels that 
he has failed, though unintentionally, to give credit for much of 
the best he has received. 

A. W. S. 
Stanford University, California, January, 1895. 



CONTENTS 



PAGE 

PREFACE V 

INTRODUCTION ix 

CHAPTER I. 

PRELIMINARY 1 

CHAPTER II. 

MOTION IN MECHANISMS . . . . . . 11 

CHAPTER III. 

ENERGY IN MACHINES .28 

CHAPTER IV. 
PARALLEL OR STRAIGHT LINE MOTIONS 37 

CHAPTER V. 

TOOTHED WHEELS, OR GEARS 39 

CHAPTER VI. 
cams . . . 72 

CHAPTER VII. 

BELTS ............ 75 

CHAPTER VIII. 

DESIGN OF FLY-WHEELS 93 

CHAPTER IX. 

RIVETED JOINTS 100 



Vlll CONTENTS. 

PAGE 

CHAPTEE X. 
DESIGN OF JOURNALS 110 

CHAPTER XI. 

SLIDING SURFACES . • 126 

CHAPTER XII. 

BOLTS AND SCREWS AS MACHINE FASTENINGS . . . .130 

CHAPTER XIII. 
MEANS FOR PREVENTING RELATIVE ROTATION . . . .137 

CHAPTER XIV. 

FORM OF PARTS AS DICTATED BY STRESS . . . .140 

CHAPTER XV. 

MACHINE SUPPORTS . . . 146 

CHAPTER XVI. 

MACHINE FRAMES . . : . ■ ; . . . . .150 

INDEX . 161 



INTRODUCTION 



In general there are four considerations of prime importance in 
designing machines: I. Adaptation, II. ■ Strength and Stiffness, 
III. Economy, IV. Appearance. 

I. This requires all complexity to be reduced to its lowest terms 
in order that the machine shall accomplish the desired result in the 
most direct way possible, and with greatest convenience to the op- 
erator. 

II. This requires the machine parts subjected to the action of 
forces to sustain these forces, not only without rupture, but also with- 
out such yielding as would interfere with the accurate action of the 
machine. In many cases the forces to be resisted may be calculated, 
and the laws of Mechanics, and the known qualities of constructive 
materials become factors in determining proportions. In other 
cases the force, by the use of a "breaking piece," may be limited to 
a maximum value, which therefore dictates the design. But in 
many other cases the forces acting are necessarily unknown ; and 
appeal must be made to the precedent of successful practice, or to 
the judgment of some experienced man, until one's own judgment 
becomes trustworthy by experience. 

In proportioning machine parts, the designer must always be 
sure that the stress which is the basis of the calculation or the esti- 
mate, is the maximum possible stress. Otherwise the part will be 
incorrectly proportioned. For instance, if the arms of a pulley 
were to be designed solely on the assumption that they endure only 
the transverse stress due to the belt tension, they would be found 
to be absurdly small, because the stresses resulting from the shrink- 



X INTRODUCTION. 

age of the casting in cooling, are often far greater than those due 
to the belt pull. 

The design of many machines is a result of what may be called 
"machine evolution." The first machine was built according to 
the best judgment of its designer ; but that judgment was fallible, 
and some part yielded under the stresses sustained ; it was replaced 
by a new part made stronger ; it yielded again, and again was en- 
larged, or perhaps made of some more suitable material ; it then 
sustained the applied stresses satisfactorily. Some other part- 
yielded too much under stress, although it was entirely safe from 
actual rupture; this part was then stiffened, and the process con- 
tinued, till the whole machine became properly proportioned for 
the resisting of stress. Many valuable lessons have been learned 
from this process ; many excellent machines have resulted from it. 
There are, however, two objections to it : it is slow and very expen- 
sive, and if any part had originally an excess of material, it is not 
changed ; only the parts that yield are perfected. 

III. The attainment of economy does not necessarily mean the 
saving of metal or labor, although it may mean that. To illustrate: 
Suppose that it is required to design an engine lathe for the market. 
The competition is sharp ; the profits are small. How shall the 
designer change the design of the lathes on the market to increase 
profits ? (a) He may, if possible, reduce the weight of metal used, 
maintaining strength and stiffness by better distribution. But this 
must not increase labor in the foundry or machine shop, nor reduce 
weight which prevents undue vibrations. (b) He may design 
special tools to reduce labor without reduction of the standard of 
workmanship. The interest on the first cost of these special tools, 
however, must not exceed the possible gain from increased profits, 
(c) He may make the lathe more convenient for the workmen. 
True economy permits some increase in cost to gain this end. 
It is not meant that elaborate and expensive devices are to be used, 
such as often come from men of more inventiveness than judgment, 
and which usually find their level in the scrap heap ; but that if the 
parts can be rearranged, or in any way changed so that the lathes- 



INTRODUCTION. XI 

man shall select this lathe to use because it is handier, when other 
lathes are available, then economy has been served, even though 
the cost has been somewhat increased ; because the favorable opin- 
ion of intelligent workmen means increased sales. 

In (a) economy is served by a reduction of metal ; in (b) by a 
reduction of labor ; in (c) it may be served by an increase of both 
labor and material. 

The addition of material largely in excess of that necessary for 
strength and rigidity, to reduce vibrations, may also be in the 
interest of economy, because it may increase the durability of the 
machine and its foundation ; may reduce the expense incident upon 
repairs and delays, thereby bettering the reputation of the machine, 
and increasing sales. 

Suppose, to illustrate further, that a machine part is to be 
designed, and either of two forms, A or B, will serve equally well. 
The part is to be of cast iron. The pattern for A will cost twice as 
much as for B. In the foundry and machine shop, however, A 
can be produced a very little cheaper than B. Clearly then, if but 
one machine is to.be built, B should be decided on ; whereas, if the 
machine is to be manufactured in large numbers, A is preferable. 
Expense for patterns is a first cost. Expense for work in the foun- 
dry and machine shop is repeated with each machine. 

Economy of operation also needs attention. This depends upon 
the efficiency of the machine ; i. e., upon the proportion of the 
energy supplied to the machine which really does useful work. This 
efficiency is increased by the reduction of useless friction a I resist- 
ances, by careful attention to the design and meang of lubrication, 
of rubbing surfaces. 

In order that economy may be best attained, the machine 
designer needs to be familiar with all the processes used in the 
construction of machines — pattern making, foundry work, forging, 
and the processes of the machine shop — and must have them con- 
stantly in mind, so that while each part designed is made strong 
enough and stiff enough, and properly and conveniently arranged, 



Xll INTRODUCTION. 

and of such form as to be satisfactory in appearance, it also is 
so designed that the cost of construction is a minimum. 

IV. The fourth important consideration is Appearance. There 
is a beauty possible of attainment in the design of machines which is 
always the outgrowth of a purpose. Otherwise expressed : A ma- 
chine to be beautiful must he purposeful. Ornament for ornament's 
sake is seldom admissible in machine design. And yet the striving 
for a pleasing effect is as much a part of the duty of the machine 
designer as it is a part of the duty of an architect. 



ELEflENTARY flACHINE DESIGN. 



CHAPTER I. 



PRELIMINARY 



1. The solution of problems in machine design involves the 
consideration of force, motion, work, and energy. It is assumed 
that the student understands clearly what is meant by these terms. 

A complete cycle of action of a machine is such an interval 
that all conditions in the machine are the same at its beginning 
and end. 

The law of Conservation of Energy underlies every machine prob- 
lem. This law may be expressed as follows : The sum of energy 
in the universe is constant. Energy may be transferred in space ; 
it may be changed from one of its several forms to another ; but it 
cannot be created or destroyed. 

The application of this law to machines is as follows : A machine 
receives energy from a source, and uses it to do useful and useless 
work. During a complete cycle of action of the machine, the 
energy received equals the total work done. In other words, a 
machine gives out, in some way, during each cycle, all the energy 
it receives ; but it cannot give out more than it receives ; or, con- 
sidering a cycle of action, 

energy received = useful work -(- useless work. 

When any two of these quantities are given, or can be esti- 
mated, the third quantity becomes known. 



Xll INTRODUCTION. 

and of such form as to be satisfactory in appearance, it also is 
so designed that the cost of construction is a minimum. 

IV. The fourth important consideration is Appearance. There 
is a beauty possible of attainment in the design of machines which is 
always the outgrowth of a purpose. Otherwise expressed : A ma- 
chine to be beautiful must be purposeful. , Ornament for ornament's 
sake is seldom admissible in machine design. And yet the striving 
for a pleasing effect is as much a part of the duty of the machine 
designer as it is a part of the duty of an architect. 



ELEflENTARY flACHINE DESIGN. 



CHAPTER I. 



PRELIMINARY. 



1. The solution of problems in machine design involves the 
consideration of force, motion, work, and energy. It is assumed 
that the student understands clearly what is meant by these terms. 

A complete cycle of action of a machine is such an interval 
that all conditions in the machine are the same at its beginning 
and end. 

The law of Conservation of Energy underlies every machine prob- 
lem. This law may be expressed as follows : The sum of energy 
in the universe is constant. Energy may be transferred in space ; 
it may be changed from one of its several forms to another ; but it 
cannot be created or destroyed. 

The application of this law to machines is as follows : A machine 
receives energy from a source, and uses it to do useful and useless 
work. During a complete cycle of action of the machine, the 
energy received equals the total work done. In other words, a 
machine gives out, in some way, during each cycle, all the energy 
it receives ; but it cannot give out more than it receives ; or, con- 
sidering a cycle of action, 

energy received = useful work + useless work. 

When any two of these quantities are given, or can be esti- 
mated, the third quantity becomes known. 



2 MACHINE DESIGN. 

2. Function of Machines. — Nature furnishes sources of energy, 
and the supplying of human needs requires tvork to be done. The 
function of machines is to cause matter possessing energy to do useful 
work. 

Illustration. — The water in a mill pond possesses energy (poten- 
tial) by virtue of its position. The earth exerts an attractive force 
upon it. If there is no outlet, the earth's attractive force cannot 
cause motion ; and hence, since motion is a necessary factor of 
work, no work is done. 

If the water overflows the dam, the earth's attraction causes that 
part of it which overflows to move to a lower level, and before it can 
be brought to rest again, it does work against the force which brings 
it to rest. If this water simply falls upon rocks, its energy is 
transformed into heat, with no useful result. 

But if the water be led from the pond to a lower level, in a 
closed pipe which connects with a water-wheel, it will exert pressure 
upon the vanes of the wheel (because of the earth's attraction), and 
will cause the wheel and its shaft to rotate against resistance, 
whereby it may do useful work. The water-wheel is a machine and 
is called a Prime Mover, because it is the first link in the machine 
chain between natural energy and useful work. 

Since it is usually necessary to do the required work at some 
distance from the necessary location of the water-wheel, Machinery 
of Transmission is used (shafts, pulleys, belts, cables, etc.), and the 
rotative energy is rendered available at the required place. 

But this rotative energy may not be adapted to do the required 
work; the rotation may be too slow or too fast; a resistance may 
need to be overcome in straight, parallel lines, or at periodical inter- 
vals. Hence Machinery of Application is introduced to transform 
the energy to meet the requirements of the work to be done. Thus 
the chain is complete, and the potential energy of the water does 
the required useful work. 

The chain of machines which has the steam boiler and engine 
for its prime mover, transforms the potential heat energy of fuel 
into useful work. This might be analyzed in a similar way. 



PRELIMINAKY. 6 

3. Force Opposed by Passive Resistance. — A force may act without 
being able to produce motion (and hence without being able to do 
work), as in the case of the water in a mill pond without overflow 
or outlet. This may be further illustrated : Suppose a force, say 
hand pressure, to be applied vertically to the top of a table. The 
material of the table offers a passive resistance, and the force is 
unable to produce motion, or to do work. But if the table top 
were supported upon springs, the applied force would overcome the 
elastic resistance of the springs, through a certain space, and would 
do work. It is therefore possible to offer passive resistance to such 
forces as may be required not to produce motion ; thereby rendering 
them incapable of doing work. 

4. Constrained Motion. — By watching the action of a machine, it 
is seen that certain definite motions occur, and that any departure 
from these motions, or the production of any other motions, would 
result in derangement of the action of the machine. Thus, the 
spindle of an engine lathe turns accurately about its axis ; the 
cutting tool moves parallel to the spindle's axis ; and an accurate 
cylindrical surface is thereby produced. If there were any depart- 
ure from these motions, the lathe would fail to do its required work. 
In all machines certain definite motions must be produced, and 
all other motions must be prevented ; or, in other words, motion in 
machines must be constrained. 

In the case of a " free body " acted on by a sj^stem of forces, not 
in equilibrium, motion results in the direction of the resultant of 
the system. If another force be introduced whose line of action does 
not coincide with that of the resultant, the direction of the line of 
action of the resultant is changed, and the body moves in this new 
direction. The character of the motion, therefore, is dependent upon 
the forces which produce the motion. This is called free motion. 

Example. — In Fig. 1, suppose the free body M to be acted on 
by the concurrent forces i, 2, and 3, whose lines of action pass 
through the center of gravity of M. The line of action of the result- 
ant of these forces is A B, and the body's centre of gravity would 
move along this line. 



4 MACHINE DESIGN. 

If another force, 4, be introduced, CD becomes the line of action 
of the resultant, and the motion of the body is along the line CD. 

Constrained motion differs from free motion in being independ- 
ent of the forces which produce it. If any force, not sufficiently 
great to produce deformation, be applied to a body whose motion is 
constrained, the result is either a certain predetermined motion, or 
no motion at all. 

In a machine there must be provision for resisting every pos- 
sible force which tends to produce any but the required motion. 
This provision is usually made by means of the passive resistance of 
properly formed and sufficiently resistant metallic surfaces. 

Illustration I. — Fig. 2 represents a section of a wood lathe 
headstock. It is required that the spindle, S, and the attached 
cone pulley, C, shall have no other motion than rotation about the 
axis of the spindle. If any other motion is possible, this machine 
part cannot be used for the required purpose. At A and B the 
cylindrical surfaces of the spindle are enclosed by accurately fitted 
bearings, or internal cylindrical surfaces. Suppose any force, P, 
whose line of action lies in the plane of the paper, to be applied 
to the cone pulley. It may be resolved into a radial component, 
R, and a tangential component, T. The passive resistance of the 
cylindrical surfaces of the journal and its bearing, prevents R from 
producing motion ; while it offers no resistance, friction being 
disregarded, to the action of T, which is allowed to produce the 
required motion, i.e., rotation about the spindle's axis. If the line 
of action of P pass through the axis, its tangential component 
becomes zero, and no motion results. If the line of action of 
P become tangential, its radial component becomes zero, and 
P is wholly applied to produce rotation. If a force Q, whose line 
of action lies in the plane of the paper, be applied to the cone, 
it may be resolved into a radial component, N, and a component, M, 
parallel to the spindle's axis. N is resisted as before by the 
journal and bearing surfaces, and M is resisted by the shoulder 
surfaces of the bearings, which fit against the shoulder surfaces of 
the cone pulley. The force Q can therefore produce no motion at all. 



PRELIMINARY. O 

In general, any force applied to the cone pulley may be resolved 
into a radial, a tangential, and an axial component. Of these only 
the tangential component is able to produce motion; and that 
motion is the motion required. The constrainment is therefore 
complete ; i. e., there can be no motion except rotation about the 
spindle's axis. This result is due to the passive resistance of 
metallic surfaces. 

Illustration II. — R, Fig. 3, represents, with all details omitted, 
the "ram," or portion of a shaping machine which carries the cut- 
ting tool. It is required to produce plane surfaces, and hence 
the "ram" must have accurate rectilinear motion parallel to HK. 
Any deviation from such motion renders the machine useless. 

Consider A. Any force which can be applied to the ram, may 
be resolved into three components : one vertical, one horizontal 
and parallel to the paper, and one perpendicular to the paper. 
The vertical component, if acting upward, is resisted by the plane 
surfaces in contact at C and D ; if acting downward, it is resisted 
by the plane surfaces in contact at E. Therefore no vertical com- 
ponent can produce motion. The horizontal component parallel to 
the paper is resisted by the plane surfaces in contact at .For G, 
according as it acts toward the right or left. The component per- 
pendicular to the paper is free to produce motion parallel to its 
line of action ; but this is the motion required. 

Any force, therefore, which has a component perpendicular to 
the paper, can produce the required motion; but no other motion. 
The constrainment is therefore complete, and the result is due to 
the passive resistance offered by metallic surfaces. 

Complete Constrainment is not always required in machines. It 
is only necessary to prevent such motions as interfere with the 
accomplishment of the desired result. 

The weight of a moving part is sometimes utilized to produce 
constrainment in one direction. Thus in a planer table, and in 
some lathe carriages, downward motion, and unallowable side 
motion, are resisted by metallic surfaces ; while upward motion is 
resisted by the weight of the moving part. 



6 MACHINE DESIGN. 

Since the motions of machine parts are independent of the forces 
producing them, it follows that the relation of such motions may be 
determined without bringing force into the consideration. 

5. Kinds of Motion in Machines. — Motion in machines may be 
very complex, but it is chiefly plane motion. 

When a body moves in such a way that any section of it 
remains in the same plane, its motion is called plane motion. All 
sections parallel to the above section must also remain, each in its 
own plane. If the plane motion is such that all points of the 
moving body remain at a constant distance from some line, AB, 
the motion is called rotation about the axis AB. Example. — A 
line shaft with attached parts. 

If all points of a body move in straight parallel paths, the 
motion of the body is called rectilinear translation. Examples. — 
Engine cross-head, ]athe carriage, planer table, shaper ram. Recti- 
linear translation may be conveniently considered as a special 
case of rotation, in which the axis of rotation is at an infinite 
distance, at right angles to the motion. 

If a body moves parallel to an axis about which it rotates, the 
body is said to have helical or screw motion. Example. — A nut 
turning upon a stationary screw. 

If all points of a body, whose motion is not plane motion, move 
so that their distances from a certain point, 0, remain constant, the 
motion is called spheric motion. This is because the points move 
in the surface of a sphere whose centre is 0. Example. — The balls 
of a fly-ball steam-engine governor, when the position of the valve 
is changing. 

6. Relative Motion. — The motion of any machine part, like all 
known motion, is relative motion. It is studied by reference to 
some other part of the same machine. Some one part of a machine 
is usually (though not necessarily) fixed ; i. e., it has no motion 
relatively to the earth. This fixed part is called the frame of the 
machine. The motion of a machine part may be referred to the 
frame, or, as often necessary, to some other part which also has 
motion relatively to the frame. 




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PRELIMINARY. / 

The kind and amount of relative motion of a machine part, 
depends upon the motions of the part to which its motion is 
referred. 

Illustration. — Fig. 4 shows a press. A is the frame; C is a 
plate which is so constrained that it may move vertically, but 
cannot rotate relatively to A. Motion of rotation is communicated 
to the screw B. The motion of B referred to A is helical motion, 
i. e.j combined rotation and translation. C, however, shares the 
translation of B, and hence there is left only rotation as the rela- 
tive motion of B and C. The motion of B referred to C is rotation. 
The motion of C referred to B is rotation. The motion of C 
referred to A is translation. 

In general, if two machine members, M and JV, move relatively 
to the frame, the relative motion of M referred to N depends 
on how much of the motion of N is shared by M. If M and N 
have the same motions relatively to the frame, they have no motion 
relatively to each other. 

Conversely, if two bodies have no relative motion, they have the 
same motion relatively to a third body. Thus in Fig. 4, if the con- 
strainment of C were such that it could share J3's rotation, as well 
as its translation, then C would have helical motion relatively to 
the frame, and no motion at all relatively to B. This is assumed 
to be self-evident. 

A rigid body is one in which the distance between elementary 
portions is constant. No body is absolutely rigid, but usually in 
machine members the departure from rigidity is so slight that it 
may be neglected. 

Many machine members, as springs, etc., are useful because of 
their lack of rigidity. 

Points in a rigid body can have no relative motion, and hence 
must all have the same motion. 

7. Instantaneous Motion, and Instantaneous Centres or Centros.— 
Points of a moving body trace more or less complex paths. If a 
point be considered as moving from one position in its path to 
another indefinitely near, its motion is called instantaneous motion. 



8 MACHINE DESIGN. 

The point is moving, for the instant, along a straight line joining 
the two indefinitely near together positions, and such a line is a 
tangent to the path. In problems which are solved by the aid of 
the conception of instantaneous motion, it is only necessary to 
know the direction of motion ; hence, for such purposes, the instan- 
taneous motion of a point is fully defined by a tangent to its path 
through the point. 

Thus in Fig. 5, if a point is moving in the path APB, when it 
occupies the position P the tangent TT represents its instantaneous 
motion. Any number of curves could be drawn tangent to TT at 
P, and any one of them would be a possible path of the point ; but 
whatever path it is following, its instantaneous motion is represented 
by TT. The instantaneous motion of a point, is therefore independ- 
ent of the form of its path. Any one of the possible paths may be 
considered as equivalent, for the instant, to a circle whose centre is 
anywhere in the normal NN. 

In general, the instantaneous motion of a point, A, is equivalent 
to rotation about some point, B, in a line through the point, A, per- 
pendicular to the direction of its instantaneous motion. 

Let the instantaneous motion of a point, A, Fig. 6, in a section 
of a moving body be given by the line TT. Then the motion is 
equivalent to rotation about some point of the line AB as a centre ; 
but it may be any point, and hence the instantaneous motion of the 
body is not determined. But if the instantaneous motion of 
another point, 0, be given by the line T x T t , this motion is equiva- 
lent to rotation about some point of CD. But the points A and C 
are points in a rigid body, and can have no relative motion, and 
must have the same motion, i. e., rotation about the same centre. 
A rotates about some point of AB, and C rotates about some point 
of CD; but they must rotate about the same point, and the only 
point which is at the same time in both lines, is their intersection, 0. 
Hence A and C, and all other points of the body, rotate for the 
instant about an axis of which is the projection ; or, in other 
words, the instantaneous motion of the body is rotation about an 
axis of which is the projection. This axis is the instantaneous 



PRELIMINARY. \) 

axis of the body's motion, and is the instantaneous centre of the 
motion of the section shown in Fig. 6. 

For the sake of brevity an instantaneous centre will be called a 
centre 

If TT and T X T X had been parallel to each other, AB and CD 
would also have been parallel, and would have intersected at in- 
finity ; in which case the body's instantaneous motion would have 
been rotation about an axis infinitely distant ; i. <?., it would have 
been translation. 

The motion of the body in Fig. 6, is of course referred to a fixed 
body, which, in this case, may be represented by the paper. The 
instantaneous motion of the body is rotation about relatively 
to the paper. Let M represent the figure, and N the fixed body 
represented by the paper. Suppose the material of M to be ex- 
tended so as to include 0. Then a pin could be put through 0, 
materially connecting M and TV, without interfering with their 
instantaneous motion. Such connection at any other point would 
interfere with the instantaneous motion. 

The centro of the relative motion of two bodies is a point, and the 
only one, at which they have no relative motion; it is a point, and the 
only one, that is common to the two bodies for the instant. 

It will be seen that the points of the figure in Fig. 6 might be 
moving in any paths, so long as those paths are tangent at the 
points to the lines representing the instantaneous motion. 

In general, centros of the relative motion of two bodies are con- 
tinually changing their position. They may, however, remain sta- 
tionary; i. e., they may become fixed centres of rotation. 

8. Loci of Centros, or Centroids. — As centros change position they 
describe curves of some kind, and these loci of centros may be 
called centroids. 

Suppose a section of any body, M, to have motion relatively to a 
section of another body, TV (fixed), in the same or parallel plane. 
Centros may be found for a series of positions, and a curve drawn 
through them on the plane of TV would be the centroid of the motion 
of M relatively to TV. If, now, M being fixed, TV moves so that the 



10 MACHINE DESIGN. 

relative motion is the same as before, the centroid of the motion of 
TV relatively to M, may be located upon the plane of M. Now, since 
the centro of the relative motion of two bodies is a point at which 
they have no relative motion, and since the points of the centroids 
become successively the centres of the relative motion, it follows 
that as the motion goes on, the centroids would roll upon each 
other without slipping. Therefore, if the centroids are drawn, and 
rolled upon each other without slipping, the bodies M and N will 
have the same relative motion as before. From this it follows that 
the relative plane motion of two bodies may be reproduced by roll- 
ing together, without slipping, the centroids of that motion. 

9. Pairs of Motion Elements. — The external and internal surfaces 
by which motion is constrained in Figs. 2 and 3 may be called 
pairs of motion elements. The pair in Fig. 2 is called a turning pair, 
and the pair in Fig. 3 is called a sliding pair. 

The helical surfaces by which a nut and screw engage with each 
other, are called a twisting pair. These three pairs of motion ele- 
ments have their surfaces in contact throughout. They are called 
lower pairs. Another class, called higher pairs, have contact only 
along elements of their surfaces. Examples. — Cams and toothed 
wheels. 



CHAPTER II. 



MOTION IN MECHANISMS. 



10. Linkages or Motion Chains ; Mechanisms. 

In Fig. 7, b is joined to c by a turning pair 
c " d " sliding " 

of " a • " turning " 

a " 6 " 

Evidently there is complete constrainment of the relative motion 
of a, b, c, and d. For, d being fixed, if any motion occurs in either 
a, b, or c, the other two must have a predetermined corresponding 
motion. 

c may represent the cross-head, h the connecting-rod, and a the 
crank of a steam engine of the ordinary type. If c were rigidly 
attached to a piston upon which the expansive force of steam 
acts toward the right, a must rotate about ad. This represents a 
machine. The members a, 6, c, and a!, may be represented for the 
study of relative motions by the diagram, Fig. 8. 

This assemblage of bodies, connected so that there is complete 
constrainment of motion, may be called a motion chain or linkage, 
and the connected bodies may be called links. The chain shown is 
a simple chain, because no link is joined to more than two others. 
If any of the links of a chain are joined to more than two 
others, the chain is a compound chain. Examples will be given later. 

When one link of a chain is fixed, i. e., when it becomes the 
standard to which the motion of the others is referred, the chain is 
called a mechanism. Fixing different links of a chain gives differ- 



12 MACHINE DESIGN. 

ent mechanisms. Thus in Fig. 8, if d be fixed, the mechanism is 
that which is used in the usual type of steam engine, as in Fig. 7. 
It is called the slider crank mechanism. 

Bat if a be fixed, the result is an entirely different mechanism ; 
for b would then rotate about the permanent centre ab, d would 
rotate about the permanent centre ad, while c would have a more 
complex motion, rotating about a constantly changing centro, 
whose path may be found. 

Fixing 6 ore would give, in each case, a different mechanism. 

11. Location of Centros. — In Fig. 8 d is fixed and it is required 
to find the centros of rotation, either permanent or instantaneous, 
of the other three links. The motion of a, relatively to the fixed 
link d, is rotation about the fixed centre ad. The motion of c 
relatively to d is translation, or rotation about a centro- cd, at infin- 
ity vertically. The link b has a point in common with a ; it is 
the centro, ab, of their relative motion. This point may be con- 
sidered as a point in a or b ; in either case it can have but one 
direction of motion. As a point in a its motion, relatively to d, 
is rotation about ad. For the instant, then, it is moving along a 
tangent to the circle through ab. But as a point in b, its direction 
of instantaneous motion must be the same, and hence its motion 
must be about some point in the line ad — ab, extended if necessary. 
Also b has a point, be, in common with c ; and by the same 
reasoning as above, be, as a point in b, rotates, for the instant, 
about some point of the vertical line through be. Now ab and be 
are points of a rigid body, and one rotates for the instant about 
some point of AB ; and the other rotates for the instant about 
some point, CD; hence both (as well as all other points of b) 
must rotate about the intersection of AB and CD. Hence bd is 
the centro of the motion of b relatively to d. 

The motion of a may be referred to c (fixed), and ac will be 
found (by reasoning like that applied to b) to lie at the intersec- 
tion of the lines EF and GH. 

The motion chain in Fig. 8, as before stated, is called the slider 
crank chain. 



MOTION IN MECHANISMS. 13 

12. Centros of the Relative Motion of Three Bodies Are Always in 
the Same Straight Line. — In Fig. 8 it will be seen that the three 
centros of any three links lie in the same straight line. Thus ad, 
ab, and bd, are the centros of the links a, b, and d. This is true of 
any other set of three links. Proof. — Consider a, b, and4- The centro 
ab as a point in a has a direction of instantaneous motion perpen- 
dicular to a line joining it to ad. As a point in b it has a direction 
of instantaneous motion perpendicular to a line joining it to bd. 
Therefore the lines ab — ad and ab — bd are both perpendicular to 
the direction of instantaneous motion of ab, and they also both pass 
through ab ; hence they must coincide, and therefore ab, ad, and bd 
must lie in the same straight line. But a, b, and d might be any 
three bodies whatever, which have relative plane motion, and the 
above reasoning would hold. Hence it may be stated : The three 
centros of any three bodies having relative plane motion, must lie in 
the same straight line. [The statement and proof of this important 
proposition is due to Prof. Kennedy.] 

13. Lever Crank Chain. Location of Centros. — Fig. 9 shows a 
chain of four links of unequal length joined to each other by turn- 
ing pairs. The centros ab, ad, cd, and be may be located at once, 
since they are the centros of turning pairs which join adjacent 
links to each other. The centros of the relative motion of b, c, and 
d are be, cd, and bd, and these must be in the same straight line. 
Hence bd is in the line B. The centros of the relative motion of 
a, b, and d, are ab, bd, and ad ; and these also must lie in the same 
straight line. Hence bd is in the line A. Being at the same time 
in A and B, it must be at their intersection. 

14. The Constrainment of Motion in a linkage is independent of 
the size of the motion elements. As long as the cylindrical surfaces 
of turning pairs have their axes unchanged, the surfaces themselves 
may be of any size whatever, and the motion is unchanged. The 
same is true of sliding and twisting pairs. 

In Fig. 10, suppose the turning pair connecting c and d to be 
enlarged so that it includes be. The link c now becomes a cylinder, 
turning in a ring attached to, and forming part of, the link b. be 



14 MACHINE DESIGN. 

becomes a' pin made fast in c and engaging with an eye at the end 
of b. The centros are the same as before the enlargement of cd, and 
hence the relative motion is the same. 

In Fig. 11, the circular portion immediately surrounding cd is 
attached to d. The link c now becomes a ring moving in a circular 
slot. This may be simplified as in Fig. 12, whence c becomes a 
curved block moving in a limited circular slot'in d. The centros 
remain as before, the relative motion is the same, and the linkage 
is essentially unchanged. 

15. If, in the slider crank mechanism, the turning pair whose 
axis is ab, be enlarged till ad is included, as in Fig. 13, the motion of 
the mechanism is unchanged, but the link a is now called an eccen- 
tric instead of a crank. This mechanism is usually used to com- 
municate motion from the main shaft of a steam engine to the 
valve. It is used because it may be put on the main shaft any- 
where, without interfering with its continuity and strength. 

The mechanism shown in Fig. 14 is called the " slotted cross- 
head mechanism." Its centros may be found from principles 
already given. 

This mechanism is often used as follows : One end of c, as E, is 
attached to a piston working in a cylinder attached to d. This 
piston is caused to reciprocate by the expansive force of steam or 
some other fluid. The other end of c is attached to another piston, 
which also works in a cylinder attached to d. This piston may 
pump water, or compress gas. The crank a is attached to a shaft, 
the projection of whose axis is ad. This shaft also carries a fly- 
wheel which insures approximately uniform rotation. 

16. Location of Centros in a Compound Mechanism.— It is required 
to find the centros of the compound linkage, Fig. 15. In any link- 
age, each link has a centro relatively to every other link ; hence, if 
the number of links = n, the number of centros = n(n — 1). But 
the centro ab is the same as ba ; i. e., each centro is double. Hence 

the number of centros to be located for any linkage = — — - — . In 

6x5 
the linkage Fig. 15, the number of centros = — - — =15. 



MOTION IN MECHANISMS. 15 

The portion above the link d is a slider crank chain, and the 
character of its motion is in no way affected by the attachment of 
the part below d. On the other hand, the lower part is a lever 
crank chain, and the character of its motion is not affected by its 
attachment to the upper part. The chain may therefore be treated 
in two parts, and the centros of each part may be located from 
what has preceded. Each part will have six centros, and twelve 
would thus be located, ad, however, is common to the two parts, and 
hence only eleven are really found. Four centros, therefore, remain 
to be located. They are be, cf, bf, and ce. To locate be, consider 
the three links a, b, and e, and it follows that be is in the line A 
passing through ab and ae ; considering b, d, and e, it follows that 
be is in the line B through bd and de. Hence be is at the intersec- 
tion of A and B. Similar methods locate the other centros. 

In general, for finding the centros of a compound linkage of six 
links, consider the linkage to be made up of two simple chains, and 
rind their centros independently of each other. Then take the two 
links whose centro is required, together with one of the links carry- 
ing three motion elements (as a, Fig. 15). The centros of these 
links locate a straight line, A, which contains the required centro. 
Then take the two links whose centro is required, together with the 
other link which carries three motion elements. A straight line, B, 
is thereby located, which contains the required centro, and the 
latter is therefore at the intersection of A and B. 

17. Velocity is rate of motion, or motion per unit time. 

Linear velocity is linear space moved through in unit time ; it 
may be expressed in any units of length and time ; as miles per 
hour, feet per minnte or per second, etc. 

Angular velocity is angular space moved through in unit time. 
In machines, angular velocity is usually expressed in revolutions 
per minute or per second. 

The linear space described by a point in a rotating body, or its 
linear velocity, is directly proportional to its radius, or its distance 
from the axis of rotation. This is true because arcs are propor- 
tional to radii. 



16 MACHINE DESIGN. 

If A and B are two points in a rotating body, and if r v and r. t 
are their radii, then the ratio of linear velocities 
_ linear veloc. A _r l 
linear veloc. B r t ' 

This is true whether the rotation is about a centre or a centro ; 
i. e.,it is true both for continuous or instantaneous rotation. Hence 
it applies to all cases of plane motion in machines ; because all 
plane motion in machines is equivalent to either continuous or 
instantaneous rotation about some point. 

To find the relation of linear velocity of two points in a machine 
member, therefore, it is only necessary to find the relation of the 
radii of the points. The latter relation can easily be found when 
the centre or centro is located. 

18. A vector is something which has magnitude and direction. 
A vector may be represented by a straight line, because the latter 
has magnitude (its length) and direction. Thus the length of a 
straight line, AB, may represent, upon some scale, the magnitude of 
some vector, and it may represent the vector's direction by being 
parallel to it, or by being perpendicular to it. For convenience 
the latter plan will here be used. The vectors to be represented are 
the linear velocities of points in mechanisms. The lines which 
represent vectors are also called vectors. 

A line which represents the linear velocity of a point, will be 
called the linear velocity vector of the point. The symbol of linear 
velocity will be VI. Thus VIA is the linear velocity of the point 
A. Also Va will be used as the symbol of angular velocity. 

If the linear velocity and radius of a point are known, the 
angular velocity, or the number of revolutions per unit time, may 
be found ; since the linear velocity -s- length of the circumference in 
which the point travels = angular velocity. 

All points of a rigid body have the same angular velocity. 

If the radii, and ratio of linear velocities of two points, in 
different machine members, are known, the ratio of the angular 
velocities of the members may be found as follows : 

Let A be a point in a member M, and B a point in a member N. 



MOTION IN MECHANISMS. 17 

i\ = radius of A; r a = radius of B. VIA and VIB represent the 

VIA 
linear velocities of A and B, whose ratio, 7777., is known. 

V IB 

Then VaA = 1 ^- and VaB = P^. 

VaA __ VIA 2*r 2 _ VIA r, _ VaM 
VaB~ 2n Vl X VIB~ VIB X r, ~ VaN' 

If M and N rotate uniformly about fixed centres, the ratio 



VaM 



VaN 

is constant. If either M or N rotates about a centro, the ratio is a 
varying one. 

19. To find the relation of linear velocity of two points in the 
same link, it is only necessary to measure the radii of the points, 
and the ratio of these radii is the ratio of the linear velocities of 
the points. 

In Fig. 16, let the smaller circle represent the path of A, the 

centre of the crank pin of a slider crank mechanism ; the link d 

being fixed. Let the larger circle represent the rim of a pulley, 

which is keyed to the same shaft as the crank. The pulley and the 

crank are then parts of the same link. The ratio of velocity of the 

VIA r 
crank pin centre and the pulley surface = „.„ = - . In this case 

r J VIB r Y 

the link rotates about a fixed centre. The same relation holds, 

however, when the link rotates about a centro. 

20. In Fig. 17, the link d is fixed and -7777- = 1 r^r • 

VI be be — bd 

By similar triangles this expression is also equal to -^ t • 

Hence, if the radius of the crank circle be taken as the vector of 
the constant linear velocity of ab, the distance cut off on the verti- 
cal through by the line of the connecting-rod (extended if 
necessary) will be the vector of the linear velocity of be. Project 
A horizontally upon be — bd, locating B. Then be — B is the 



18 MACHINE DESIGN. 

vector of VI of the slider, and may be used as an ordinate of the 
linear velocity diagram of the slider. By repeating the above 
construction for a series of positions, the ordinates representing the 
VI of be for different positions of the slider may be found. A 
smooth curve through the extremities of these ordinates is the 
velocity curve, from which the Vis for all points of the slider's 
stroke may be read. The scale of velocities, or the linear velocity 
represented by one inch of ordinate, equals the constant linear 
velocity of ab divided by — ab in inches. 

21. It is required to find VI of be during a cycle of action of the 
mechanism shown in Fig. 18, d being fixed, and VI of ab being con- 
stant. The two points, ab and be, may both be considered in the 
link b. All points in b move about bd relatively to the fixed link. 

TX VI ab ab — bd 

Hence TWc^^^bd- 

But a line, as MN, drawn parallel to b cuts off on the radii portions 
which are proportiontal to the radii themselves, and hence propor- 
tional to the Vis of the points. Hence 

VI ab ab — M 



Vibe be — N' 



The arc in which be moves may be divided into any number of 
parts, and the corresponding positions of ab may be located. A 
circle through M. about ad, may be drawn, and the constant radial 
distance ab — M may represent the constant velocity of ab. 
Through M t , M 2 , etc., draw lines parallel to the corresponding 
positions of b, and these lines will cut off on the corresponding line 
of c a distance which represents VI of be. Through the points thus 
determined the velocity diagram may be drawn, and the VI of be 
for a complete cycle is determined. The scale of velocities is found 
as in § 20. 

22. The relation of linear velocity of points not in the same 
link may also be found. 



MOTION IN MECHANISMS. 19 

_ . ..riot a *ty'1 h . ■ t t , 

Required „ . /line centro ab is a point common to a and 

ft, the two links considered ; d is the fixed link. Consider ab as a 
point in a ; and its VI is to that of A as their radii or distances from 
ad. Draw a vector triangle with its sides parallel to the triangle 
formed by joining A, ab, and ad. Then if the side A 1 represent the 
VI of A, the side ab x will represent the VI of ab. Consider ab as a 
point in b, and its VI is to that of B as their radii, or distances to 
bd. Upon the vector ab x , draw a triangle whose sides are parallel 
to those of a triangle formed by joining ab, bd, and B. Then, from 
similar triangles, the side B 1 is the vector of i?'s linear velocity. 

w VI of A vector A x 

CG VI oiB^ vector B, ' 

The path of B during a complete cycle may be traced, and the 
VI for a series of points may be found, by the above method ; then 
the vectors may be laid off on normals to the path through the 
points ; the velocity curve may be drawn ; and the velocity of B at 
all points becomes known. 

23. The diagram of VI of the slider of the slider crank mechan- 
ism, Fig. 17, is unsymmetrical with respect to a vertical axis through 
its centre. This is due to the angularity of the connecting-rod, and 
may be explained as follows : 

In Fig. 20, A is one angular position of the crank, and B is the 
corresponding angular position on the other side of the vertical 
through the centre of rotation. The corresponding positions of the 
slider are as shown. But for position A, the line of the connecting- 
rod, C, cuts off on the vertical through 0, a vector Oa, which 
represents the slider's velocity. For position B the vector of the 
slider's velocity is Ob. Obviously this difference is due to the 
angularity of the connecting-rod. 

In a mechanism which is equivalent to the slider crank with 
the connecting-rod always horizontal (as the slotted cross-head) 
the line of the connecting-rod would cut off on OF the same vector 



20 MACHINE DESIGN. 

for position A and position B. Hence the velocity diagram for the 
slotted cross-head mechanism is symmetrical with respect to both 
vertical and horizontal axes through its centre. In fact, if the 
crank radius (= length of link a) be taken as the vector of the VI 
of ab, the linear velocity diagram of the slider becomes a circle 
whose radius = the length of the link a. Hence the crank circle 
itself serves for the linear velocity diagram, the horizontal diameter 
representing the path of the slider. 

24. During a portion of the cycle of the slider crank mechanism, 
the slider's VI is greater than that of ab. This is also due to the 
angularity of the connecting-rod, and may be explained as follows : 
In Fig. 21, as the crank moves up from the position x, it will reach 
such a position, A, that the line of the connecting-rod extended will 
pass through B. OB in this position is the vector of the linear 
velocity of both ab and the slider, and hence their linear velocities 
are equal. When ab reaches B, the line of the connecting-rod passes 
through B; and again the vectors — and hence the linear velocities — 
of ab and the slider are equal. For all positions between A and B, 
the line of the connecting-rod will cut OB outside of the crank 
circle ; and hence the linear velocity of the slider will be greater 
than that of ab. Obviously, the linear velocity of the slider is 
greatest when the angle between crank and connecting-rod = 90°. 
This result is due to the angularity of the connecting-rod, because 
if the latter remained always horizontal, its line could never cut OB 
outside the circle. It follows that in the slotted cross-head mechan- 
ism the maximum VI of the slider = the constant VI of ab. The 
angular space BOA, Fig. 21, throughout which VI of the slider is 
greater than the VI of ab, increases with increase of angularity of 
the connecting-rod ; i. e., it increases with the ratio 

length of crank 
length of connecting-rod' 

25. A slider in a mechanism often carries a cutting tool, which 
cuts during its motion in one direction, and is idle during the return 



MOTION IN MECHANISMS. 21 

stroke. Sometimes the slider carries the piece to be cut, and the 
cutting occurs while it passes under a tool made fast to the fixed 
link, the return stroke being idle. 

The velocity of cutting is limited. If the limiting velocity be 
exceeded, the tool becomes so hot that its temper is drawn, and it 
becomes unfit for cutting. The limit of cutting velocity depends on 
the nature of the material to be cut. Thus annealed tool-steel, and 
the scale surface of cast iron, may be cut at 20 feet per minute; 
wrought iron and soft steel at 25 to 30 feet per minute ; while 
brass and the softer alloys may be cut at 40 or more feet per 
minute. There is no limit of this kind, however, to the velocity 
during the idle stroke ; and it is desirable to make it as great as 
possible, in order to increase the product of the machine. This 
leads to the design and use of "quick return" mechanisms. 

26. If, in a slider crank mechanism, the centre of rotation of the 
crank be moved, so that the line of the slider's motion does not pass 
through it, the slider will have a quick return motion. 

In Fig. 22, when the slider is in its extreme position at the 
right, the crank-pin centre is at D. When the slider is at B, the 
crank-pin centre is at C. If rotation is as indicated by the arrow, 
then while the slider moves from B to A, the crank-pin centre 
moves from C over to D. And while the slider returns from A to 
B, the crank-pin centre moves under from D to C. If the VI of the 
crank-pin centre be assumed constant, the time occupied in moving 
from D to C is less than that from C to D. Hence, the time occu- 
pied by the slider in moving from B to A is greater than that 
occupied in moving from A to B. The mean velocity during the 
forward stroke is therefore less than during the return stroke. Or 
the slider has a " quick return " motion. 

It is required to design a mechanism of this kind for a length of 
stroke = BA and for a ratio 

mean VI forward stroke _ 5 
mean VI return stroke 7 

The mean velocity of either stroke is inversely proportional to the 



22 MACHINE DESIGN. 

time occupied, and the time is proportional to the corresponding 
angle described by the crank. Hence 

mean velocity forward 5 angle /3 

mean velocity return 7 angle a " 

It is therefore necessary to divide 360° into two parts which are to 
each other as 5 to 7. Hence a = 210° and £=150°. Obviously 
#=180° — /?=30°. Place the 30° angle of a drawing triangle so 
that its sides pass through B and A. This condition may be ful- 
filled and yet the vertex of the triangle may occupy an indefinite 
number of positions. By trial may be located so that the crank 
shall not interfere with the line of the slider. being located, 
tentatively, it is necessary to find the corresponding lengths of 
crank a, and connecting-rod b. When the crank-pin centre is at 
D, AO =^b — a ; when it is at C, BO = b -f a. AO and BO are meas- 
urable values of length ; hence a and b may be found, the crank 
circle may be drawn, and the velocity diagrams may be constructed 
as in Fig. 17; remembering that the distance cut off upon a vertical 
through 0, by the line of the connecting-rod, is the vector of the VI 
of the slider for the corresponding position, when the VI of the crank- 
pin centre is represented by the crank radius. 

The construction may be checked by finding the mean heights of 

\ the velocity diagram^pr the areas, which are proportional to the 

V mean heights, above and below the horizontal line ; these should 

be to each other as 5:7. The areas may be found by use of a 

planimeter ; and these areas, divided by the length of stroke, equal 

the mean heights. 

It is required to make the maximum velocity of the forward 
stroke of the slider = 20 feet per minute, and to find the correspond- 
ing number of revolutions per minute of the crank. The maximum 
linear velocity vector of the forward stroke = the maximum height 
of the upper part of the velocity diagram ; call it V v Call the 
linear velocity vector of the crank-pin centre V 2 = crank radius. 
Let x = linear velocity of the crank-pin centre. Then 



MOTION IN MECHANISMS. 23 

V 1 _ 20 ft. per minute 



20 ft. per minute X I 
or x = 



x is therefore expressed in known terms. If now x, the space the 
crank-pin centre is required to move through per minute, be divided 
by the space moved through per revolution, the result will equal the 
number of revolutions per minute = N ; 

N= 



2- X length of crank' 



27. Fig. 23 shows a compound mechanism. The link d is the 
supporting frame, and a rotates about ad in the direction indicated, 
communicating motion to c through the slider 6, so that c vibrates 
about cd. The link e, connected to c by a turning pair at ce , causes 
/ to slide horizontally on another part of the frame or fixed link d. 
The centre of the crank-pin, ab, is given a constant linear velocity, 
and the slider,/, has motion toward the left with a certain mean 
velocity, and returns toward the right with a greater mean velocity. 
This is true because the slider / moves toward the left, while a 
moves through the angle « ; and toward the right while a moves 
through the angle ft. But the motion of a is uniform, and hence 
the angular movement a represents more time than the angular 
movement ft ; and/, therefore, has more time to move toward the 
left, than it has to move through the same space toward the right. 
It therefore has a "quick return" motion. 

The machine is driven so that the crank-pin centre moves uni- 
formly, and the velocity, at all points of its stroke, of the slider 
carrying a cutting tool, is required. The problem, therefore, is to find 
the relation of linear velocities of ef and ab, for a series of positions 
during the cycle ; and to draw the diagram of velocity of ef. 

Solution. — ab has a constant known linear velocity. The point 
in the link c which coincides, for the instant, with ab, receives 
motion from <ib ; but the direction of its motion is different from 



24 MACHINE DESIGN. 

that of ab, because ab rotates about ad while the coinciding point of 
c rotates about cd. If ab-A be laid off representing the linear 
velocity of ab, then ab-B will represent the linear velocity of 
the coinciding point of the link c. Let the latter point be 
called x. 

Locate cf, at the intersection of e with the line cd — ad. Now 
cf and x are both points in the link c, and hence their linear veloci- 
ties, relatively to the fixed link, are proportional to their distances 
from cd. These two distances may be measured directly, and with 
the known value of linear velocity of x = ab-B, give three known 
values of a simple proportion, from which the fourth term, the 
linear velocity of cf, may be found. 

Or, if the line BD be drawn parallel to cd-ad, the triangle B-D-ab 
is similar to the triangle cd-cf-ab, and from the similarity of these 
triangles, it follows that BD represents the linear velocity of cf on 
the same scale that ab-B represents the linear velocity of x. Hence 
the linear velocity of cf, for the assumed position of the mechanism, 
becomes known. But since cf is a point of the slider, all of whose 
points have the same linear velocity, it follows that the linear 
velocity of cf is the required linear velocity of the slider. 

This solution may be made for as many positions of the mech- 
anism as are necessary to locate accurately the velocity curve. 

Having drawn the velocity diagram, suppose that it is required 
to make the maximum linear velocity of the slider on the slow 
stroke = A feet per minute. Then the linear velocity of the crank- 

, pin centre ab = y ={A ' _ ' — =- J If r = the crank radius, 

the number of revolutions per minute = ^—. 

When this mechanism is embodied in a machine, a becomes a 
crank attached to a shaft whose axis is at ad. The shaft turns in 
bearings provided in the machine frame. The crank carries a pin 
whose axis is at ab, and this pin turns in a bearing in the sliding 
block b. The link c becomes a lever keyed to a shaft whose axis is 
at cd. This lever has a long slot in which the block b slides. The 



MOTION IN MECHANISMS. 25 

link e becomes a connecting-rod, connected both to c and / by pin 
and bearing. The link/ becomes the "cutter bar'' or "ram" of a 
shaper : the part which carries the cutting tool. The link d 
becomes the frame of the machine, which not only affords support 
to the shafts at ad and cd, and the guiding surfaces for/, but also 
is so designed as to afford means for holding the pieces to be planed, 
and supports the feeding mechanism. 

28. Fig. 24 shows another compound linkage, d is fixed, and c 
rotates uniformly about cd, communicating rotary motion to a 
through the slider b. a is extended past ad (the part extended 
being in another parallel plane), and moves a slider/ through a 
link e. This is called the " Whitworth quick return mechanism." 
The point be at which c communicates motion to a moves along a, 
and hence the radius of the point at which a receives a constant 
linear velocity varies, and the angular velocity of a must vary 
inversely. Hence the angular velocity of a is a maximum when 
the radius is a minimum, i. e., when a and c are vertical down- 
ward ; and the angular velocity of a is minimum when the radius 
is a maximum, i. e., when a and c are vertical upward. 

29. Problem. — To design a Whitworth Quick Return for a given 

mean VI of f forward 
ratio, 



mean VI of/ returning' 



When the centre of the crank-pin, C, reaches A, the point D will 
coincide with B< the link C will occupy the angular position cd-B, 
and the slider /will be at its extreme position toward the left. 

When the point C reaches F, the point D will coincide with E, 
the link e will occupy the angular position cd-E, and the slider / 
will be at its extreme position toward the right. 

Obviously, while the link c moves over from the position cd-E to 
the position cd-B, the slider / will complete its forward stroke. 
While c moves under from cd-B to cd-E,f will complete the return 
stroke. The link c moves with a uniform angular velocity, and 
hence the mean velocity of / forward is inversely proportional to 
the angle /?, and the mean velocity of/ returning is inversely pro- 



26 MACHINE DESIGN. ' 

mean VI of /forward 



portional to «. Or 

v mean VI of/ returning p 

For the design, the distance cd-ad must be known. This may 

usually be decided on from the limiting sizes of the journals at cd 

a 5 
and ad. Suppose that the above ratio ==-==•=, that cd-ad = 3", 

and that the maximum length of stroke of /= 12". Locate cd 
and measure off vertically downward a distance equal to 3", thus 
locating ad. Draw a horizontal line through ad. The point ef of 
the slider/ will move along this line. 

' a 5 
Since -=■=-=•, and a 4-/3 = 360°, 

.-. ar=150°and ,3 = 210°. 
Lay off « from cd as a centre, so that the vertical line through cd 
bisects it. Draw a circle through B with cd as a centre, B being 
the point of intersection of the bounding line of « with a horizontal 
through ad. The length of the link c = cd-B. 

The radius ad-C must equal the travel of /-j-2 = 6". This 
radius is made adjustable, so that the length of stroke may be 
varied. The connecting-rod, e, may be made of any convenient 
length. 

30. Problem. — To draw the velocity diagram of the slider /of 
the Whitworth Quick Return. The point be, Fig. 25, has a known 
constant linear velocity, and its direction of motion is always at 
right angles to a line joining it to cd. The point of the link a, 
which coincides in this position of the mechanism with be, receives 
motion from be, but its direction of motion is at right angles to the 
line bc-ad. If be- A represent the linear velocity of be, its projection 
upon bc-ad extended will represent the linear velocity of the point 
of a which coincides with be. Call this point x. The centro af 
may be considered as a point in a, and its linear velocity relatively 
to d, when so considered, is proportional to its distance from ad. 

tj VI of af ad-af 

VI of x ad-bc ' 



MOTION IN MECHANISMS. 27 



But the triangles ad-api-bc, and B-C-bc are similar. Hence 

VI of af _ BC 
VI of x ~ B-bc ' 

This means that BC represents the linear velocity of af upon the 
same scale that B-bc represents the linear velocity of x. But af is a 
point in /, and all points in / have the same linear velocity ; hence 
BC represents the linear velocity of the slider /, for the given posi- 
tion of the mechanism, and it may be laid off as an ordinate of the 
velocity curve. This solution may be made for as many positions 
as are required to locate accurately the entire velocity curve for a 
cycle of the mechanism. 



CHAPTER III. 



ENERGY IN. MACHINES. 



31. The subject of motion and velocity, in certain simple 
machines, has been treated and illustrated. It remains now to 
consider the passage of energy through similar machines. From 
this the solution of force problems will follow. 

During the passage of energy through a machine, or chain of 
machines, any one. or all, of four changes may occur. 

I. The energy may be transferred in space. Example. — Energy 
is received at one end of a shaft and transferred to the other end, 
where it is received and utilized by a machine. 

II. The energy may be converted into another form. Examples. 
— (a) Heat energy into mechanical energy by the steam engine 
machine chain, (b) Mechanical energy into heat by friction. 
(c) Mechanical energy into electrical energy, as in a dynamo- 
electric machine ; or electrical energy into mechanical energy in 
the electric motor, etc. 

III. Energy is the product of a force factor and a space factor. 
Energy per unit time, or rate of doing work, is the product of a 
force factor and a velocity factor ; since velocity is space per unit 
time. Either factor may be changed at the expense of the other ; 
i. e., velocity may be changed, if accompanied by such a change of 
force that the energy per unit time remains constant. Correspond- 
ingly, force may be changed at the expense of velocity, energy per 
unit time being constant. Example. — A belt transmits 6000 foot- 
pounds of energy per minute to a machine. The belt velocity is 



ENERGY IN MACHINES. 29 

120 feet per minute, and the force exerted is 50 lbs. Frictional 
resistance is neglected. A cutting tool in the machine does useful 
work ; its velocity is 20 feet per minute, and the resistance to cut- 
ting is 300 lbs. Then, energy received per minute = 120 X 50 = 
6000 foot-pounds ; and energy delivered per minute = 20 X 300 = 
6000 foot-pounds. The energy received therefore equals the energy 
delivered. But the velocity and force factors are quite different in 
the two cases. 

IV. Energy may be transferred in time. In many machines 
the energy received at every instant equals that delivered. There 
are many cases, however, where there is a periodical demand for 
work, i. e., a fluctuation in the rate of doing work ; while energy 
can only be supplied at the average rate. Or there may be a uni- 
form rate of doing work, and a fluctuating rate of supplying energy. 
In such cases means are provided in the machine, or chain of 
machines, for the storing of energy till it is needed. In other words, 
energy in transferred in time. Examples. — (a) In the steam engine 
there is a varying rate of supplying energy during each stroke, 
while there is (in general) a uniform rate of doing work. There is, 
therefore, a periodical excess and deficiency of effort. A heavy 
wheel on the main shaft absorbs the excess of energy with increased 
velocity, and gives it out again with reduced velocity, when the 
effort is deficient, (b) A pump delivers water into a pipe system 
under pressure. The water is used in a hydraulic press, whose 
action is periodic, and beyond the capacity of the pump. A 
hydraulic accumulator is attached to the pipe system, and while 
the press is idle the pump slowly raises the accumulator weight, 
thereby storing potential energy, which is given out rapidly by the 
descending weight for a short time while the press acts, (c) A 
dynamo-electric machine is run by a steam engine, and the electri- 
cal energy is delivered and stored in storage batteries, upon which 
there is a periodical demand. In this case, as well as in case (b), 
there is a transformation of energy as well as a transfer in time. 

32. Force Problems. — Suppose the slider crank mechanism in 
Fig. 26 to represent a shaping machine ; the velocity diagram of 



30 MACHINE DESIGN. 

the slider being drawn. The resistance offered to cutting metal 
during the forward stroke must be overcome. This resistance may 
be assumed constant. Throughout the cutting stroke there is a 
constantly varying rate of doing work. This is because the rate of 
doing work = resisting force (constant) X velocity (varying). This 
product is constantly varying, and is a maximum when the slider's 
velocity is a maximum. The slider must be driven by means of 
energy transmitted through the crank a. The maximum rate at 
which energy must be supplied, equals the maximum rate of doing 
work at the slider. Draw the mechanism in the position of max- 
imum velocity of slider ; i. e., locate the centre of the slider-pin at 
the base of the maximum ordinate of the velocity diagram, and 
draw b and a in their corresponding positions. The slider's known 
velocity is represented by y, and the crank-pin's required velocity 
is represented by z on the same scale. Hence the value of^rbecomes 
known by simple proportion. The rate of doing work must be the 
same at c and at ab (neglecting friction).* Hence Rv 1 = Fv 2 , in which 
R and i\ represent the force and velocity factors at c ; and F and v. 2 
represent the force and velocity factors at ab. R and v t are known 
from the conditions of the problem, and v 2 is found as above. 

Rv 
Hence, F may be found, = — x = force which, applied tangentially 

to the crank-pin centre, will overcome the maximum resistance of 
the machine. In all other positions of the cutting stroke the rate 
of doing work is less, and F would be less. But it is necessary to 
provide driving mechanism capable of overcoming the maximum 
resistance, when no fly-wheel is used. If now F be multiplied by 
the crank radius, the product equals the maximum torsional 
moment ( = M) required to drive the machine. If the energy is 
received on some different radius, as in case of gear or belt trans- 
mission, the maximum driving force = M^- the new radius. During 
the return stroke the cutting tool is idle, and it is only necessary to 

*The effect of acceleration to redistribute energy is zero in this position, 
because the acceleration of the slider at maximum velocity is zero, and the 
angular acceleration of b can only produce pressure in the journal at ad. 



£&.&. 




r/a.26. 



ENERGY IN MACHINES. 31 

overcome the frictional resistance to motion of the bearing surfaces. 
Hence, the return stroke is not considered in designing the driving 
mechanism. When the method of driving this machine is decided 
on, the capacity of the driving mechanism must be such that it 
shall be capable of supplying to the crank shaft the torsional 
driving moment M, determined as above. 

This method applies as well to the quick return mechanisms 
given. In each, when the velocity diagram is drawn, the vector of 
the maximum linear velocity of the slider, — L x , and of the constant 
linear velocity of the crank-pin centre, = L 2 , are known, and the 
velocities corresponding, V x and V 2 , are also known, from the scale 
of velocities. The rate of doing work at the slider and at the crank- 
pin centre is the same, friction being neglected. Hence Rv x = Fv 2 , 
or, since the vector lengths are proportional to the velocities they 

represent, RL X = FL 2 ; and F= -=- - . Therefore the resistance to the 

slider's motion =R, on the cutting stroke, multiplied by the ratio of 

linear velocity vectors, y 1 , of slider and crank-pin, equals F, the 

maximum force that must be applied tangentially at the crank- 
pin centre to insure motion. F multiplied by the crank radius 
= maximum torsional driving moment required by the crank 
shaft. If R is varying, and known, find where Rv, the rate of 
doing work, is a maximum, and solve for that position in the 
same way as above. 

33. Force Problems — Continued. — In the usual type of steam 
engine the slider crank mechanism is used, but energy is supplied 
to the slider (which represents piston, piston-rod, and cross-head), 
and the resistance opposes the rotation of the crank and attached 
shaft. In any position of the mechanism, Fig. 28, force applied to 
the crank-pin through the connecting-rod, may be resolved into 
two components, one radial and one tangential. The tangential 
component tends to produce rotation ; the radial component pro- 
duces pressure between the surfaces of the shaft journal and its 
bearing. The tangential component is a maximum when the angle 



82 MACHINE DESIGN. 

between crank and connecting-rod equals 90°; and it becomes zero 
when C reaches A or B. If there is a uniform resistance, the rate of 
doing work is constant. Hence, since the energy is supplied at a 
varying rate, it follows that during part of the revolution the effort 
is greater than the resistance ; while during the remaining portion 
of the revolution the effort is less than the resistance. When effort 
is less than resistance, the machine will stop unless other means 
are provided to maintain motion. A "fly-wheel" is keyed to the 
shaft, and this wheel, because of slight variations of velocity, alter- 
nately stores and gives out the excess and deficiency of energy of 
the effort, thereby adapting it to the constant work to be done. 

34. Problem. — Given length of stroke of the slider of a steam 
engine slider crank mechanism, the required horse-power, or rate of 
doing work, and number of revolutions. Required the total mean 
pressure that must be applied to the piston. 

Let L = length of stroke = 1 foot ; 

HP = horse-power = 20 ; 
N = r ^oluxio ^*per minute = 200 ; 
F == required mean force on piston. 

Then NXL— 200 feet per minute = mean velocity of slider = V. 

Now, the mean rate of doing work in the cylinder and at the main 

shaft during each stroke is the same (friction neglected); hence FV 

= HP X 33000, 

„ HP X 33000 20X33000 _ onn _, 
F-= y- -^-- = 3300 lbs. 

35. In the slider crank chain, the velocity of the slider neces- 
sarily varies from zero, at the ends of its stroke, to a maximum 
value near mid-stroke. The mass of the slider and attached parts 
is therefore positively and negatively accelerated each stroke. 
When a mass is positively accelerated it stores energy ; and when 
it is negatively accelerated it gives out energy. The amount of this 
energy, stored or given out, depends upon the mass and the accel- 
eration. The slider stores energy during the first part of its stroke, 
and gives it out during the second part of its stroke. While, there- 



n$.2& 




a 



^cn 



ENEKGY IN MACHINES. 33 

fore, it gives out all the energy it receives, it gives it out differently 
distributed. In order to find exactly how the energy is distributed, 
it is necessary to find the acceleration throughout the slider's 
stroke. This may be done as follows : Fig. 27 A shows the velo- 
city diagram of the slider of a slider crank mechanism, for the 

forward stroke. The acceleration required at any point = -7—, in 

which dv is the increase in velocity during any interval of time 
M, assuming that the increase in velocity becomes constant at that 
point. Lay off the horizontal line OP=MN. Divide OP into as 
many equal parts as there are unequal parts in MN. These divis- 
ions may each represent M. At m erect the ordinate mn = m 1 n l ; 
and at o erect the ordinate op = o l p v Continue this construction 
throughout OP, and draw a curve through the upper extremities of 
the ordinates. B is a velocity diagram on a "time base." At 
draw the tangent OT to the curve. If the increase in velocity were 
constant during the time interval represented by Om, the 
increment of velocity would be represented by mT. Therefore 
???T'is proportional to the acceleration at the point 0, and may be 
laid off as an ordinate of an acceleration diagram C. Thus Qa — 
mT. The divisions of QR are the same as those of MN; i. e., they 
represent positions of the slider. This construction may be repeated 
for the other divisions of the curve B. Thus at n the tangent nT 1 
and horizontal nq are drawn, and qT x is proportional to the accel- 
eration at n, and is laid off as an ordinate be of the acceleration 
diagram. To find the value in acceleration units of Qa, mT is read 
off in velocity units = 4v, by the scale of ordinates of the velocity 
diagram. This value is divided by 4t, the time increment corres- 

ponding to Om. The result of this division -p = acceleration at 

M in acceleration units. ^£ = the time of one stroke, or of one- 
half revolution of the crank, divided by the number of divisions in 
OP. If the linear velocity of the centre of the crank-pin, = V, be 

MN 
represented by the length of the crank radius — -=—= r, then the 

5 ^ 



34 MACHINE DESIGN. 

scale of velocities, or velocity in feet per second for 1" of ordinate 

V rtDN 

= — = — — . D is the actual diameter of the crank circle, and r is 
r r oO 

the crank radius measured on the figure. If the weight, W, of parts 
accelerated is known, the force, F, necessary to produce the accel- 
eration at any slider position may be found from the fundamental 
formula of mechanics 

9 
p being the acceleration corresponding to the position considered. 
If the ordinates of the acceleration diagram be taken as represent- 
ing the forces which produce the acceleration, the diagram will have 
force ordinates and space abscissae, and areas will represent work. 
Thus, Qas represents the work stored during acceleration, and Rsd 
represents the work given out during retardation. Let MN, Fig. 29, 
represent the length of the slider's stroke, and NC the resistance of 
cutting (uniform); then energy to do cutting per stroke is repres- 
ented by the area MBCN. But during the early part of the stroke 
the reciprocating parts must be accelerated, and the force necessary 
at the beginning, found as above, = BD. The driving gear must, 
therefore, be able to overcome resistance equal to MB + BD. The 
acceleration, and hence the accelerating force, decreases as the 
slider advances, becoming zero at E. From E on the acceleration 
becomes negative, and hence the slider gives out energy and helps 
to overcome the resistance, and the driving gear has only to furnish 
energy represented by the area AEFN, though the work really 
done against resistance equals that represented by the area CEFN. 
The energy represented by the difference of these areas, = ACE, is 
that which is stored in the slider's mass during acceleration. Since 
by the law of conservation of energy, energy given out per cycle = 
that received, it follows that area ACE = &rea, DEB, and area 
BCMN=ADMN. This redistribution of energy would seem to 
modify the problem on page 30, since that problem is based on the 
assumption of uniform resistance during cutting stroke. The 
position of maximum velocity of slider, however, corresponds to 



ENERGY IN MACHINES. 35 

acceleration =0. The maximum rate of doing work, and the cor- 
responding torsional driving moment at the crank shaft would 
prohably correspond to the same position, and would not be materi- 
ally changed. In such machines as shapers, the acceleration and 
weight of slider are so small that the redistribution of energy is 
unimportant. 

36. Solution of the force problem in the steam engine slider crank 
mechanism ; slider represents piston with its rod, and the cross- 
head. — The steam acts upon the piston with a pressure which 
varies during the stroke. The pressure is redistributed before 
reaching the cross-head pin, because the reciprocating parts are 
accelerated in the first part of the stroke, with accompanying 
storing of energy and reduction of. pressure on the cross-head 
pin ; and retarded in the second part of the stroke, with accom- 
panying giving out of energy and increase of pressure on the 
cross-head pin. Let the ordinates of the full line diagram above 
OX, Fig. 30 A, represent the effective pressure on the piston 
throughout a stroke. B is the velocity diagram of slider. Find 
the acceleration throughout stroke, and from this and the known 
value of weight of slider, find the force due to acceleration. Draw 
diagram 0, whose ordinates represent the force due to accelera- 
tion, upon the same force scale used in A. Lay off this diagram 
on OX as a base line, thereby locating the dotted line. The 
vertical ordinates between this dotted line and the upper line of 
A represent the pressure applied to the cross-head pin. These 
ordinates may be laid off from a horizontal base line, giving 
D. The product of the values of the corresponding ordinates 
of B and D = the rate of doing work throughout the stroke. Thus 
the value of GH in pounds X value of EF in feet per second 
= the rate of doing work in foot-pounds per second upon the 
cross-head pin, when the centre of the cross-head pin is at E. 
The rate of doing work at the crank-pin is the same as at the cross- 
head pin. Hence dividing this rate of doing work, = EF X GH, 
by the constant tangential velocity of the crank-pin centre, gives 
the force acting tangentially on the crank-pin to produce rota- 



36 MACHINE DESIGN. 

tion. The tangential forces acting throughout a half revolution 
of the crank may be thus found, and plotted upon a horizontal 
base line = length of half the crank circle, Fig. 31. The work done 
upon the piston, cross-head pin, and crank during a piston stroke is 
the same. Hence the areas of A and D, Fig. 30, are equal to 
each other, and to the area of the diagram, Fig. 31. The forces 
acting along the connecting-rod for all positions during the piston 
stroke, may be found by drawing force triangles with one side hor- 
izontal, one vertical, and one parallel to position of connecting-rod 
axis, the horizontal side being equal to the corresponding ordinate 
of D. The vertical sides of these triangles will represent the guide 
reaction, while the side parallel to the connecting-rod axis repre- 
sents the force transmitted by the connecting-rod. 



ce 

o— 



kc c \A 



hat, 



U> 



D. 



C&3L 




F/g.32. 



«mo 



7777p| 



4 



CHAPTER IV. 

PARALLEL OR STRAIGHT LINE MOTIONS. 

37. Rectilinear motion in machines is usually obtained by 
means of prismatic guides. It is sometimes necessary, however, to 
accomplish the same result by linkages. 

A general method of design, which is applicable in many cases, 
will be given. In Fig. 32, d is the fixed link, and a is connected 
with it by a sliding pair, a, 6, c, and e are connected by turning 
pairs, as shown. The constrainment is not complete because B is 
free to move in any direction, and its motion would, therefore, 
depend upon the force producing it. It is required that the point 
B shall move in a straight line parallel to a. Suppose that B is 
caused to move along the required line ; then any point of the link 
c, as A, will describe some curve, FAE. If a pin be attached to c, 
with its axis at A, and a curved slot fitting the pin, with its sides 
parallel to FAE, be attached to d, as in Fig. 33, it follows that B 
can only move in the required straight line. This is the mechanism 
of the Tabor Steam Engine Indicator. 

The curve described by A might approximate a circular arc whose 
centre could be located, say at 0, Fig. 33. Then the curved slot 
might be replaced by a link, attached to d and c by turning pairs 
at and A. This gives B approximately the required motion. 
This is the mechanism of the Thompson Steam Engine Indicator. 

If, while the point B is caused to move in the required straight 
line, a point in b, as P, Fig. 32, were chosen, it would be found to 
describe a curve which would approximate a circular arc, whose 
centre, 0, and radius, = r, could be found. Let the link whose 



38 MACHINE DESIGN. 

length = r be attached to d and b by turning pairs whose axes are 
at and P, and the motion of B will be approximately the 
required motion. This is the mechanism of the Crosby Steam 
Engine Indicator. 

38. Problem. — In Fig. 34, B is the fixed axis of a counter-shaft ; 
C is the axis of another shaft which is free to rotate about B. D is 
the axis of a circular saw which is free to move in any direction. 
It is required to constrain D to move in the straight line EF. If D 
be moved along EF, a tracing point fixed at A in the link CD will 
describe an approximate circular arc, HAK, whose centre may be 
found at 0. A link whose length is OA may be connected to the 
fixed link, and to the link CD by means of turning pairs at and 
A. D will then be constrained to move approximately along EF. 
A curved slot and pin could be used, and the motion would be 
exact.* 

* Descriptions of many varieties of parallel motions may be found in Kan- 
kine's " Machinery and Millwork " ; Weisbach's " Mechanics of Engineering," 
Vol. Ill, ''Mechanics of the Machinery of Transmission"; Kennedy's 
" Mechanics of Machinery." 



CHAPTER V. 

TOOTHED WHEELS, OK GEARS. 

39. When toothed wheels are used to communicate motion, the 
motion elements are the tooth surfaces. The contact of these sur- 
faces with each other is line contact. Such pairs of motion elements 
are called higher pairs, to distinguish them from lower pairs, which 
are in contact throughout their entire surface. Fig. 35 shows the 
simplest toothed wheel mechanism. There are three links, a, b, and 
c, and therefore three centros ab, be, and ac. These centros must, as 
heretofore explained, lie in the same straight line, ac and ab are 
the centres of the turning pairs connecting c and b to a. It is re- 
quired to locate be on the line of centres. 

When the gear c is caused to rotate uniformly with a certain 
angular velocity, i. e., at the rate of m revolutions per minute, it is 
required to cause the gear b to rotate uniformly at a rate of n revo- 
lutions per minute. The angular velocity- ratio is therefore con- 

TYl 

stant, and = — . The centro be is a point on the line of centres 
n r 

which has the same linear velocity whether it is considered as a 

point in b or e. The linear velocity of this point be in b = 2^R l n ; 

and the linear velocity of the same point in c = 27zR. z m; in which 

R } = radius of be in b, and R 2 = radius of be in c. But this linear 

velocity must be the same in both cases, and hence the above 

expressions may be equated thus : 

2xR{n = 2nR t m ; 

, R, m 

whence — 1 = — . 

R., n 



40 MACHINE DESIGN. 

Hence be is located by dividing the line of centres into parts which 

are to each other inversely as the angular velocities of the gears. 

Thus, let ab and ac, Fig. 36, be the centres of a pair of gears 

771 

whose angular velocity ratio = — . Draw the line of centres ; divide 

. n 

into 777 + 7i equal parts ; m of these from ab toward the right, or n 
from ac toward the left, will locate be. Draw circles through be, 
with ab and ac as centres. These circles are the centroids of be and 
are called pitch circles. It has been already explained that any 
motion may be reproduced by rolling the centroids of that motion 
upon each other without slipping. Therefore the motion of gears 
is the same as that which would result from the rolling together of 
the pitch circles (or cylinders) without slipping. In fact, these 
pitch cylinders themselves might be, and sometimes are, used for 
transmitting motion of rotation. Slipping, however, is apt to occur, 
and hence these "friction gears" cannot be used if no variation 
from the given velocity ratio is allowable. Hence, teeth are formed 
on the wheels which engage with each other, to prevent slipping. 

40. Teeth of almost any form may be used, and the average 
velocity will be right. But if the forms are not correct there will 
be continual variations of velocity ratio between a minimum and 
maximum value. These variations are in many cases unallowable, 
and in all cases undesirable. It is necessary therefore to study 
tooth outlines which shall serve for the transmission of a constant 
velocity ratio. 

The centro of relative motion of the two gears must remain in a 
constant position in order that the velocity ratio shall be constant. 
The essential condition for constant velocity ratio is, therefore, that the 
position of the centro of relative motion of the gears shall remain un- 
changed. If A and B, Fig. 3ftf are tooth surfaces in contact at a, 
their only possible relative motion, if they remain in contact, is 
slipping motion along the tangent CD. The centro of this motion 
must be in EF, a normal to the tooth surfaces at the point of con- 
tact. If these be supposed to be teeth of a pair of gears, b and c, 
whose required velocity ratio is known, and whose centro, be, is 



TOOTHED WHEELS, OR GEARS. 41 

therefore located, then in order that the motion communicated from 
one gear to the other through the point of contact, a, shall be the 
required motion, it is necessary that the centro of the relative motion 
of the teeth shall coincide with be. 

Illustration. — In Fig. 38, let ac and ab be centres of rotation 
of bodies b and c, and the required velocity ratio is such that the 
centro of b and c falls at be. Contact between b and c is at p. The 
only possible relative motion if these surfaces remain in contact is 
slipping along CD ; hence the centro of this motion must be on EF, 
the normal to the tooth surfaces at the point of contact. But it must 
also be on the same straight line with ac and ab ; hence it is at be, 
and the motion transmitted for the instant, at the point p, is the 
required motion, because its centro is at be. But the curves touch- 
ing at p, might be of such form that their common normal at p 
would intersect the line of centres at some other point, as K, which 
would then become the centro of the motion of b and c for the 
instant, and would correspond to the transmission of a different 
motion. The essential condition to be fulfilled by tooth outlines, in 
order that a constant velocity ratio may be maintained, may there- 
fore be stated as follows : The tooth outlines must be such that their 
normal at the point of contact shall always pass through the centro 
corresponding to the required velocity ratio. 

41. Having given any curve that will serve for a tooth outline 

in one gear, the corresponding curve may be found in the other 

gear, which will engage with the given curve and transmit a con- 

m 
stant velocity ratio. Let — be the given velocity ratio. Draw the 

line of centres AB, Fig. 39. Let P be the " pitch point," i. e., the 
point of contact of the pitch circles or the centro of relative motion 
of the two gears. To the right from P lay off a distance PB — m ; 
from P toward the left lay off PA =n. A and B will then be the 
required centres of the wheels, and the pitch circles may be drawn 
through P. Let abc be any given curve on the wheel A. It is 
required to find the curve in B which shall engage with abc to 
transmit the constant velocity ratio required. A normal to the 



42 MACHINE DESIGN. 

point of contact must pass through the centro. If, therefore, any 
point, as a, be taken in the given curve, and a normal to the curve 
at that point be drawn, as a«, then when a is the point of contact, 
a will coincide with P. Also, if cy is a normal to the curve at c, 
then y will coincide with P when c is the point of contact between 
the gears ; and since b is in the pitch line, it will itself coincide 
with P when it is the point of contact. Suppose now that A and B 
are discs of cardboard, that A overlaps B, and that a thread is 
stretched to indicate the centre line AB. Suppose also that they 
can be rotated so that the pitch circles roll on each other without 
slipping. Roll the discs till a reaches P, and prick a through upon 
B ; then make b coincide with P, and prick it through ; then make 
y coincide with P, and prick c through. This will give three points 
in the required curve in B, and through these the curve may be 
drawn. The curve could, of course, be more accurately located by 
using more points. The points of the curve in B might be located 
geometrically. 

Many curves could be drawn that would not serve for tooth out- 
lines ; but, given any curve that will serve, the corresponding curve 
may be found. There would be, therefore, almost an infinite 
number of curves, that would fulfill the requirements of correct 
tooth outlines. But in practice two kinds of curves are found so 
convenient that they are most commonly, though not exclusively, 
used. They are cycloidal and involute curves. 

42. It is assumed that the character of cycloidal curves and 
method of drawing them is understood. 

In Fig. 40, let b and c be the pitch circles of a pair of wheels, 
always in contact at be. Also let m be the describing circle in con- 
tact with both at the same point. M is the describing point. 
When one curve rolls upon another, the centro of their relative 
motion is always their point of contact. For, since the motion of 
rolling excludes slipping, the two bodies must be stationary, rela- 
tively to each other, at their point of contact ; and bodies that 
move relatively to each other can have but one such stationary 
point in common — their centro. When, therefore, m rolls in or 



TOOTHED WHEELS, OR GEARS. 43 

upon b or c, its centro relatively to either is their point of contact. 
The point M, therefore, must describe curves whose direction at 
any point is at right angles to a line joining that point to the point 
of contact of m with the circle. Suppose the two circles b and c 
to revolve about their Centres, being always in contact at be ; sup- 
pose m to rotate at the same time, the three circles being always in 
contact at one point. The point M will then describe simultane- 
ously a curve, 6', on the plane of b, and a curve, c', on the plane of c. 
Since M describes the curves simultaneously, it will always be the 
point of contact between them in any position. And since the 
point M moves always at right angles to a line which joins it to be, 
therefore the normal to the tooth surfaces at their point of contact 
will always pass through be, and the condition for constant velocity 
ratio transmission is fulfilled. But these curves are precisely the 
epicycloid and hypocycloid that would be drawn by the point M in 
the generating circle, by rolling on the outside of b and inside of c. 
Obviously, then, the epicycloids and hypocycloids generated in this 
way, used as tooth profiles, will transmit a constant velocity ratio. 

This proof is independent of the size of the generating circle, and 
its diameter may therefore equal the radius of b. Then the hypo- 
cycloids generated by rolling within b would be straight lines coin- 
ciding with the radius of b. In this case the profiles of the teeth of 
b become radial lines ; and therefore the teeth are thinner at the 
base than at the pitch line ; for this reason they are weaker than if 
a smaller generating circle had been used. All tooth curves gener- 
ated with the same generating circle will work together, the pitch 
being the same. It is therefore necessary to use the same generat- 
ing circle for a set of gears which need to interchange. 

The describing circle may be made still larger. In the first case 
the curves described have their convexity in the same direction, 
i. e., they lie on the same side of a common tangent. When the 
diameter of the describing circle is made equal to the radius of b, 
one curve becomes a straight line tangent to the other curve. As 
the describing circle becomes still larger, the curves have their 
convexity in opposite directions. As the circle approximates 



44 MACHINE DESIGN. 

equality with b, the hypocycloid grows shorter, and finally, when 
the describing circle equals b, it becomes a point which is the gener- 
ating point in b, which is now the generating circle. If this point 
could be replaced by a pin having no sensible diameter, it would 
engage with the epicycloid generated by it in the other gear to 
transmit a constant velocity ratio. But a pin without sensible 
diameter will not serve as a wheel tooth, and a proper diameter 
must be assumed, and a new curve laid off to engage with it in the 
other gear. In Fig. 41, AB is the epicycloid generated by a point 
in the circumference of the other pitch circle. CD is the new curve 
drawn tangent to a series of positions of the pin as shown. The 
pin will engage with this curve, CE, and transmit the constant 
velocity ratio as required. In Fig. 40, let it be supposed that when 
the three circles rotate constantly tangent to each other at the pitch 
point be, a pencil is fastened at the point M in the circumference of 
the describing circle. If this pencil be supposed to mark simul- 
taneously upon the planes of b, c, and that of the paper, it will 
describe upon b an epicycloid, on c a hypocycloid, and on the plane 
of the paper an arc of the describing circle. Since M is always 
the point of contact of the cycloidal curves (because it generates 
them simultaneously), therefore, in cycloidal gear teeth, the locus 
or path of the point of contact is an arc of the describing circle. 

43. In the cases already considered, where an epicycloid in one 
wheel engages with a hypocycloid in the other, the contact of the 
teeth with each other is all on one side of the line of centres. Thus, 
in Fig. 40, if the motion be reversed, the curves will be in contact 
until M returns to be along the arc MD-bc ; but after M passes be 
contact will cease. If c were the driving wheel, the point of con- 
tact would approach the line of centres ; if b were the driving wheel 
the point of contact would Tecede from the line of centres. Exper- 
ience shows that the latter gives smoother running because of better 
conditions as regards friction between the tooth surfaces. It would 
be desirable, therefore, that the wheel with the epicycloidal curves 
should always be the driver. But it should be possible to use either 
wheel as driver to meet the varying conditions in practice. 



TOOTHED WHEELS, OR GEARS. 45 

Another reason why contact should not be all on one side of the 
line of centres may be explained as follows : 

Definitions. — The angle through which a gear wheel turns, while 
one of its teeth is in contact with the corresponding tooth in the 
other gear, is called the angle of action. The arc of the pitch circle 
corresponding to the angle of action is called the arc of action. 

The arc of action must be greater than the "pitch arc" (the arc 
of the pitch circle that includes one tooth and one space), or else 
contact will cease between one pair of teeth before it begins between 
the next pair. Constrainment would therefore not be complete. 

In Fig. 42, let AB and CD be the pitch circles of a pair of gears, 
and E the describing circle. Let an arc of action be laid off on 
each of the circles from P, as Pa, Pc, and Pe. Through e, about the 
centre 0, draw an addendum circle, i. e., the circle which limits the 
points of the teeth. Since the circle E is the path of the point of 
contact, and since the addendum circle limits the points of the 
teeth, their intersection, e, is the point at which contact ceases, 
rotation being as indicated by the arrow. If the pitch arc just 
equals the assumed arc of action, contact will be just beginning at 
/' when it ceases at e ; but if the pitch arc be greater than the arc 
of action, contact will not begin at P till after it has ceased at e, 
and there will be an interval when AB will not drive CD. The 
greater the arc of action the greater the distance of e from P on 
the circumference of the describing circle. The direction of pressure 
between the teeth is always a normal to the tooth surface, and this 
always passes through the pitch point ; therefore, the greater the arc of 
action , i.e., the greater the distance of e from P, the greater the obliquity 
of the line of pressure. The pressure may be resolved into two com- 
ponents, one at right angles to the line of centres, and the other 
parallel to it. The first is resisted by the teeth of the follower 
wheel, and therefore produces rotation ; the second is resisted at the 
journal, and produces pressure, with resulting friction. Hence, it 
follows that the greater the arc of action, the greater will be the 
average obliquity of the line of pressure, and therefore the greater 
the component of the pressure that produces wasteful friction. If 



46 MACHINE DESIGN. 

it can be arranged so that the arc of action shall be partly on each 
side of the line of centres, the arc of action may be made greater 
than the pitch arc (usually equal to about 1-J times the pitch 
arc); then the obliquity of the pressure line may be kept within 
reasonable limits, contact between the teeth will be insured in all 
positions, and either wheel may be the driver. This is accomplished 
by using two describing circles as in Fig. 43. ■ Suppose the four 
circles A, B, «, and ft, to roll constantly tangent at P. a will de- 
scribe an epicycloid on the plane of B, and a hypocycloid on the 
plane of A. These curves will engage with each other to drive 
correctly, ft will describe an epicycloid on A, and a hypocycloid 
on B. These curves will engage also, to drive correctly. If the 
epi- and hypocycloid in each gear be drawn through the same point 
on the pitch circlej a double curve tooth outline will be located, 
and one curve will engage on one side of the line of centres, and the 
other on the other side. If A drives as indicated by the arrow, 
contact will begin at D, and the point of contact will follow an arc 
of a to P, and then an arc of ft to C. 

44. Involute Tooth Outlines. — If a string be wound around a 
cylinder and a pencil point attached to its end, this point will 
trace an involute, as the string is unwound from the cylinder. Or, 
if the point be constrained to follow a tangent to the cylinder, and 
the string be unwound by rotating the cylinder about its axis, the 
point will trace an involute on a plane that rotates with the cylin- 
der. Illustration. — Let a, Fig. 44, be a circular piece of wood, free 
to rotate about ; ft is a circular piece of cardboard made fast to 
a ; AB is a straight-edge held on the circumference of a, having a 
pencil point at B. As B moves along the straight-edge to A, « and 
ft rotate about C, and B traces an involute DB upon ft. The rela- 
tive motion of the tracing point and ft being just the same as if the 
string had been simply unwound from a, fixed. If the tracing 
point is caused to return along the straight-edge it will trace the 
involute BD in a reverse direction. 

The centro of the tracing point is always the point of tangency 
of the string with the cylinder ; therefore the string, or straight- 



TOOTHED WHEELS, OR GEARS. 47 

edge, in Fig. 4M is always at right angles to the direction of motion 
of the tracing point, and hence is always a normal to the involute curve. 
Let « and ft, Fig. 45, he two base cylinders ; let AB be a cord wound 
upon a and ft and passing through the centro P, which corresponds 
to the required velocity ratio. Let « arid ft be supposed to rotate so 
that the cord is wound from ft upon a. Then any point in the cord 
will move from A toward B, and, if it be a tracing point, will trace 
an involute of ft on the plane of ft (extended beyond the base cylin- 
der), and will also trace an involute of a upon the plane of «. 
These two involutes will serve for tooth profiles for the transmission 
of the required constant velocity ratio, because AB is the constant 
normal to both curves at their point of contact, and it passes 
through P, the centro corresponding to the required velocity ratio. 
Hence, the necessary condition is fulfilled. 

Since a point in the line AB describes involute curves simul- 
taneously, the point of contact of the curves is always in the line 
AB. And hence AB is the path of the point of contact. 

One of the advantages of involute curves for tooth profiles is 
that a change in distance between centres of the gears, does not 
interfere with the transmission of a constant velocity ratio. This 
may be proved as follows : In Fig. 45, from similar triangles 

7^r-r = 7^7r; that is, the ratio of the radii of the base circles is 
A OP 

equal to the ratio of the radii of the pitch circles. This ratio 
equals the inverse ratio of angular velocities of the gears. Suppose 
now that and 0' be moved nearer together : the pitch circles will 
be smaller, but the ratio of their radii must be equal to the un- 
changed ratio of the radii of the base circles, and therefore the 
velocity ratio remains unchanged. Also the involute curves, since 
they are generated from the same base cylinders, will be the same 
as before, and therefore, with the same tooth outlines, the same 
constant velocity ratio will be transmitted as before. 

45. Definitions. — If the pitch circle be divided into as many 
equal parts as there are teeth in the gear, the arc included between 
two of these divisions is the circular pitch of the gear. Circular 



48 MACHINE DESIGN. 

pitch may also be defined as the distance on the pitch circle occu- 
pied by a tooth and a space ; or, otherwise, it is the distance on the 
pitch circle from any point of a tooth to the corresponding point in 
the next tooth. A fractional tooth is impossible, and therefore the 
circular pitch must be such a value that the pitch circumference is 
divisible by it. Let P = circular pitch in inches; let D = pitch 

diameter in inches ; N — number of teeth ; then NP = tD ; N=— ; 

NP 7rD 

D= ; P = -rr- From these relations any one of the three 

values, P, D, and N, may be found if the other two are given. 

Diametral pitch is the number of teeth per inch of pitch diameter. 

N 
Thus, if p = diameter pitch, p— — . Multiplying the two expres- 

sions, r= -?f and p — — , togetlier, gives Pp = -~- . — = ?r. (Jr, 

the product of diametral and circular pitch =w. Circular pitch is 
usually used for large cast gears, and for mortice gears (gears with 
wooden teeth inserted). Diametral pitch is usually used for small 
cut gears. 

In Fig. 46, b, e, and k, are pitch points of the teeth ; ab is the 
face of the tooth ; bm is the flank of the tooth ; AD is the total depth 
of the tooth ; AC is the working depth; AB is the addendum; a 
circle through A is the addendum circle. Clearance is the excess of 
total depth over working depth, = CD. Backlash is the width of 
space on the pitch line, minus the width of the tooth on the same 
line. In cast gears whose tooth surfaces are not " tooled " backlash 
needs to be allowed, because of unavoidable imperfections in the 
surfaces. In cut gears, however, it may be reduced almost to zero, 
and the tooth and space, measured on the pitch circle, may be con- 
sidered equal. 

46. Racks. — A rack is a wheel whose pitch radius is infinite ; 
its pitch circle, therefore, becomes a straight line, and is tangent 
to the pitch circle of the wheel, or pinion with which the rack 
engages. The line of centres is a normal to the pitch line of the 




Vj/ js&az. 



TOOTHED WHEELS, OR GEARS. 49 

rack, through the centre of the pitch circle of the pinion. The 
pitch of the rack is determined hy laying off the circular pitch of 
the engaging wheel on the pitch line of the rack. The curves of 
the rack teeth, like those of wheels of finite radius, may he generated 
by a point in the circumference of a circle which rolls on the pitch 
circle. Since, however, the pitch circle is now a straight line, the 
tooth curves will be cycloids, both for flanks and faces. In Fig. 47, 
AB is the pitch circle of the pinion, and CD is the pitch line of the 
rack ; a and b are describing circles. Suppose, as before, that all 
move without slipping, and are constantly tangent at P. A point 
in the circumference of a will then describe simultaneously a 
cycloid on CD, and a hypocycloid within AB. These will be 
correct tooth outlines. Also, a point in the circumference of h 
will describe a cycloid on CD and an epicycloid on AB. These 
will be correct tooth outlines. To find the path of the point 
of contact, draw the addendum circle EF of the pinion, and the 
addendum line GH of the rack. When the pinion turns clockwise 
and drives the rack, contact will begin at e and follow arcs of the 
describing circles through P to K. It is obvious that a rack cannot 
be used where rotation is continuous in one direction, but only 
where motion is reversed. 

Involute curves may also be used for the outlines of rack teeth. 
Let CD and CD', Fig. 48, be the pitch lines. When it is required 
to generate involute curves for tooth outlines, for a pair of gears of 
finite radius, a line is drawn through the pitch point at a given 
angle to the line of centres (usually 75°); this line is the path of 
the point which generates two involutes simultaneously, and there- 
fore the path of the point of contact between the tooth curves. It 
is also the common tangent to the two base circles, which may now 
be drawn and used for the describing of the involutes. To apply 
this to the case of a rack and pinion, draw EF, Fig. 48. The base 
circles must be drawn tangent to this line ; AB will therefore be 
the base circle for the pinion. But the base circle in the rack has 
an infinite radius, and a circle of infinite radius drawn tangent to 
EF would be a straight line coincident with EF. Therefore EF is 



50 MACHINE DESIGN. 

the base line of the rack. But an involute to a base circle of 
infinite radius is a straight line normal to the circumference — in 
this case a straight line perpendicular to EF. Therefore the tooth 
profiles of a rack in the involute system will always be straight 
lines perpendicular to the path of the describing point, and passing 
through the pitch points. If, in Fig. 48, the pinion move clockwise 
and drive the rack, the contact will begin at t E, the intersection of 
the addendum line of the rack GH and the base circle AB of the 
pinion, and will follow the line EF through P to the point where 
EF cuts the addendum circle LM of the pinion. 

47. Annular Gears. — Both centres of a pair of gears may be on 
the same side of the pitch point. This arrangement corresponds to 
what is known as an annular gear and pinion. Thus, in Fig. 49, 
AB and CD are the pitch circles, and their centres are both above 
the pitch point P. Teeth may be constructed to transmit rotation 
between AB and CD. AB will be an ordinary spur pinion, but it 
is obvious that CD becomes a ring of metal with teeth on the inside, 
i. e., it is an annular gear. In this case a and ft may be describing 
circles, and a point in the circumference of « will describe hypo- 
cycloids simultaneously on the planes of AB and CD ; and a point 
in the circumference of ft will describe epicycloids simultaneously on 
the planes of AB and CD. These will engage to transmit a con- 
stant velocity ratio. Obviously the space inside of an annular gear 
corresponds to a spur gear of the same pitch and pitch diameter, 
with tooth curves drawn with the same describing circle. Let EF 
and GH, Fig. 49, be the addendum circles. If the pinion move 
clockwise, driving the annular gear, the path of the point of contact 
will be from e along the circumference of a to P, and from P along 
the circumference of ft to K. 

The construction of involute teeth for an annular gear and 
pinion involves exactly the same principle as in the case of a pair 
of spur gears. The only difference of detail is that the describ- 
ing point is in the tangent to the base circles produced instead of 
being between the points of tangency. Let and 0', Fig. 50, be 
the centres, and AB and // the pitch circles of an annular gear 



c — ^ 




h 



C6£A 



FHD 



F/a. 40. 







A 



TOOTHED WHEELS, OR GEARS. 51 

and pinion. Through P, the point of tangency of the pitch circles, 
draw the path of the point of contact, at the given angle with the 
line of centres. With and 0' as centres draw tangent circles to this 
line. These will be the involute base circles. Let the tangent be 
replaced by a cord, made fast say at K' , winding on the circumfer- 
ence of the base circle CK', to D, and then around the base circle 
FE in the direction of the arrow, and passing over the pulley G 
which holds it in line with PB. If rotation be supposed to occur 
with the two pitch circles always tangent at P without slipping, 
any point in the cord beyond P toward G, will describe an involute 
on the plane //, and another on the plane of AB. These will be 
the correct involute tooth profiles required. Draw NQ and LM, the 
addendum circles. Then if the pinion move clockwise, driving the 
annular gear, the point of contact starts from e and moves along 
the line GH through P to K. 

When a pair of spur gears mesh with each other, the direction of 
rotation is reversed. But an annular gear and pinion meshing 
together, rotate in the same direction. 

48. Interchangeable Sets of Gears. — In practice it is desirable to 
have "interchangeable sets" of gears ; i. e., sets in which any gear 
will "mesh" correctly with any other, from the smallest pinion to 
the rack, and in which, except for limiting conditions of size, any 
spur gear will mesh with any annular gear. Interchangeable sets 
may be made in either the cycloidal or involute system. A neces- 
sary condition in any set is, that the pitch shall be constant; be- 
cause the thickness of tooth on the pitch line must always equal 
the width of the space (less clearance). If this condition is 
unfulfilled they cannot engage, whatever the form of the tooth 
outlines. 

The second condition for an interchangeable set in the cycloidal 
system is that the size of the describing circle shall be constant. If 
the diameter of the describing circle equal the radius of the smallest 
pinion's pitch circle, the flanks of this pinion's teeth will be radial 
lines, and the tooth will therefore be thinner at the base than at 
the pitch line. As the gears increase in size with this constant size 



52 MACHINE DESIGN. 

of describing circle, the teeth grow thicker at the base ; hence, the 
weakest teeth are those of the smallest pinion. 

It is found unadvisable to make a pinion with less than twelve 
teeth. If the radius of a fifteen -tooth pinion be selected for the 
diameter of the describing circle, the flanks in a twelve-tooth pinion 
will be very nearly parallel, and may therefore be cut with a mill- 
ing cutter. This would not be possible if the describing circle 
were made larger, causing the space to become wider at the bot- 
tom than at the pitch circle. Therefore the maximum describ- 
ing circle for milled gears is one whose diameter equals the 
pitch radius of a rifteen-tooth pinion, and it is the one usually 
selected. Each change in the number of teeth with constant pitch 
causes a change in the size of the pitch circle. Hence, the form of 
the tooth outline, generated by a describing circle of constant 
diameter, also changes. For any pitch, therefore, a separate cutter 
would be required corresponding to every number of teeth, to in- 
sure absolute accuracy. Practically, however, this is not necessary. 
The change in the form of tooth outline is much greater in a small 
gear, for any increase in the number of teeth, than in a large one. 
It is found that twenty-four cutters will cut all possible gears of 
any pitch with sufficient practical accuracy. The range of these 
cutters is indicated in the following table, taken from Brown & 
Sharpe's "Treatise on Gearing" : 



itte 


:r A cuts 12 teeth. 


" 


B 


a 


13 


a 


a 


C 


a 


14 


u 


a 


D 


a 


15 


a 


a 


E 


a 


16 


a 


a 


F 


a 


17 


a 


u 


G 


a 


18 


u 


a 


H 


a 


19 


u 


u 


I 


U 


20 


a 


a 


J 


a 


21 to 22 teeth 


a 


K 


a 


23 to 24 " 


a 


L 


a 


24 to 26 " 



itte 


r M cuts 


27 to 


29 teeth. 


a 


N 


u 


30 to 


33 


a 


u 





iC 


34 to 


37 


a 


a 


P 


a 


38 to 


42 


u 


u 


Q 


a 


43 to 


49 


a 


a 


R 


a 


50 to 


59 


a 


u 


S 


a 


60 to 


74 


a 


a 


T 


u 


75 to 


99 


u 


u 


U 


a 


100 to 149 


a 


u 


V 


u 


150 to 249 


a 


a 


W 


a 


250 to rack. 




a 


X 


a 


rack. 







TOOTHED WHEELS, OR GEARS. 53 

These same principles of interchangeable sets of gears, with 
cycloidal tooth outlines, apply not only to small milled gears as 
above, but also to large cast gears with tooled or untooled tooth 
surfaces. 

49. Interchangeable Involute Gears. — In the involute system the 
second condition of interchangeability is that the angle between the 
common tangent to the base circles and the line of centres shall be con- 
stant. This may be shown as follows : Draw the line of centres, 
AB, Fig. 51. Through P, the assumed pitch point, draw CD, and 
let it be the constant common tangent to all base circles from which 
involute tooth curves are to be drawn. Draw any pair of pitch 
circles tangent at P, with their centres in the line AB. About 
these centres draw circles tangent to CD ; these are base circles, 
and CD may represent a cord that winds from one upon the other. 
A point in this cord will generate, simultaneously, involutes that 
will engage for the transmission of a constant velocity ratio. But 
this is true of any pair of circles that have their centres in AB, and 
are tangent to CD. Therefore, if the pitch is constant, any pair of 
gears that have the base circles tangent to the line CD, will mesh 
together properly. As in the cycloidal gears, the involute tooth 
curves vary with a variation in the number of teeth, and, for abso- 
lute theoretical accuracy, there would be required for each pitch as 
many cutters as there are gears with different numbers of teeth. 
The variation is least at the pitch line, and increases with the dis- 
tance from it. The involute teeth are usually used for the finer 
pitches, and the cycloidal teeth for the coarser pitches ; and since 
the amount that the tooth surface extends beyond the pitch line in- 
creases with the pitch, it follows that the variation in form of tooth 
curves is greater in the coarse pitch cycloidal gears than in the fine 
pitch involute gears. For this reason, with involute gears, it is 
only necessary to use eight cutters for each pitch. The range is 
shown in the following table, which is also taken from Brown & 
Sharpens " Treatise on Gearing " . 



54 



MACHINE DESIGN, 



55 ' 


' to 


134 


35 ' 


' to 


54 


26 ' 


1 to 


34 


21 ' 


' to 


25 


17 ' 


' to 


20 


14 ' 


' to 


16 


12 ' 


' to 


13 



No. 1 will cut wheels from 135 teeth to racks. 



" 4 

" 5 

" 6 

u 7 

" 8 
50. Laying Out G-ear Teeth. Exact and Approximate Methods. — 

There is ordinarily no reason why an exact method for laying out 
Cycloidal or involute curves for tooth outlines should not be used, 
either for large gears or gear patterns, or in making drawings. It is 
required to lay out a cycloidal gear. The pitch, and diameters of 
pitch, and describing circle are given. — Draw the pitch circle. From 
a piece of thin wood cut out a template to fit a segment of the pitch 
circle from the inside, as A, Fig. 52. Cut another template to fit a 
segment of the pitch circle from the outside, as B. Also cut a 
wooden disc whose diameter equals that of the given describing 
circle, and fix a tracing point in its circumference. Divide the 
pitch circle into parts equal to the given circular pitch. Let P be 
one of the pitch points. Locate A so that its curved edge coincides 
with the pitch circle at the right of P. Roll the describing circle 
on A, without slipping, so that the epicycloid described by the 
tracing point shall pass through P. Next place B so that its curved 
edge coincides with the pitch circle at the left of P, and roll the 
circle on the inside of B, without slipping, so that the hypocycloid 
described by the tracing point shall pass through P. Thus the 
outline of one tooth is drawn, aPb. Cut a wooden template to fit 
the tooth curve, and make it fast to a wooden arm free to rotate 
about 0, making the edge of the template coincide with aPb. It 
may now be swung successively to the other pitch points, and the 
tooth outline may be drawn by the template edge. This gives one 
side of all of the teeth. The arm may now be turned over and the 
other sides of the teeth may be drawn similarly. 



TOOTHED WHEELS, OR GEARS. 55 

51. It is required to lay out exact involute teeth. The pitch, 
pitch circle diameter, and angle of the common tangent are given. 
— Draw the pitch circle, Fig. 53, and the line of centres AB. 
Through the pitch point, P, draw CD, the common tangent to the 
base circles, making the angle ft with the line of centres. Draw the 
base circle about 0, tangent to CD. Cut a wooden template to fit 
the base circle from the inside, as EF ; wind on this template a 
fine cord carrying a pencil at its end, and then unwind this, allow- 
ing the pencil to trace an involute curve, ab, which will be a correct 
tooth form. Let a template, cut to fit this involute, be attached to 
an arm free to rotate about 0, and the tooth outlines may be drawn 
as before. The bottom of the spaces between the teeth may fall 
within the base circle, in which case the involute curves are extended 
inward by radial lines.* 

52. The following formulas are given to assist in the practical 
proportioning of gears : 

Let D = pitch diameter. 
" Dj= outside diameter. 

" D t = diameter of a circle through the bottom of spaces. 
" P = circular pitch = space on the pitch circle occupied by a 

tooth and a space. 
" p = diametral pitch = number of teeth per inch of pitch 

circle diameter. 
" N= number of teeth. 
" t = thickness of tooth on pitch line. 
" . a = addendum. 
" c = clearance. 
" d = working depth of spaces. 
" d 1 = full depth of spaces. 

* Approximate tooth outlines may be drawn by the use of instruments, 
such as the Willis odontograph, which locates the centres of approximate cir- 
cular arcs; the templet odontograph, invented by Prof. S. W. Robinson; or 
by some geometrical or tabular method for the location of the centres of ap- 
proximate circular arcs. For descriptions see "Elements of Mechanism," 
Willis; "Kinematics," McCord; "Teeth of Gears," George B. Grant; 
"Treatise on Gearing," published by Brown & Sharpe. 



56 MACHINE DESIGN. 

Then, D,= N + 2 ; B, == D — 2(a + c); 

JV = -p ; P = -^ ; D = — ; N= Dp; 

N n N t> 
* = D'' D= ? ** = *> 

_ _ p _ 

» = -= ; P = — ; ( = - = r- , no backlash. 
r P p 2 2p 

c = ro = £ = 5o ; d==2o; d .= 2a + c ; «= £ inches - 

The following dimensions are given as a guide ; they may be 
varied as conditions of design require : Width of face = about SP ; 
thickness of rim = 1.25 t; thickness of arms = 1.25 £ ; no taper. 
The rim may be reinforced by a rib, as shown in Fig. 54. Diame- 
ter of hub = 2 X diameter of shaft. Length of hub = width of 
face -f- \" ; width of arm at junction with hub = £ circumference 
of the hub, for six arms. Make arms taper about £" per foot on 
each side. 

53. Strength of (rear Teeth. — The maximum work transmitted 
by a shaft per unit time may usually be accurately estimated ; and, 
if the rate of rotation is known, the torsional moment may be found. 
Let 0, Fig. 55, represent the axis of a shaft perpendicular to the 
paper. Let A = maximum work to be transmitted per minute : 
let N = revolutions per minute ; let Fr = torsional moment. Then 
F is the force factor of the work transmitted, and 2*rN is the space 
factor of the work transmitted. Hence, 2F*rN = A, and Fr = tor- 
sional moment = -p— =- T . 

If the work is to be transmitted to another shaft by means of a 
spur gear whose radius is r„ then for equilibrium F^, = Fr, and 

Fr 

F, == — . Fj is the force at the pitch surface of the gear whose 

radius is r 19 i. e., it is the force to be sustained by the gear teeth. 
Hence, in general, the force sustained by the teeth of a gear equals 
the torsional moment divided by the pitch radius of the gear. 



TOOTHED WHEELS, OR GEARS. 57 

When the maximum force to be sustained is known the teeth 
may be given proper proportions. The dimensions upon which the 
tooth depends for strength are : Thickness of tooth = t : width of 
face of gear = b ; and depth of space between teeth — I. These all 
become known when the pitch is known, because t is fixed for any 
pitch, and I and b have values dictated by good practice. The 
value of b may be varied through quite a range to meet the require- 
ments of any special case. 

54. In the design the tooth will be treated as a cantilever with a 
load applied at its end. It is assumed that one tooth sustains the 
entire load ; i. e., that there is contact only between one pair of 
teeth. This would be nearly true for gears with low numbers of 
teeth ; but in high numbered gears the force would be distributed 
over several pairs, and hence they would have an excess of strength. 
It is also assumed that the tooth has the same thickness from the 
pitch circle to its root. This is also nearly true for low numbered 
gears, while high numbered gears would have excess of strength as 
a result of this assumption. In Fig. 56 let P = force at the pitch 
surface of the gear to be designed, b = width of face, I — depth of 
space, and d = thickness of the tooth at the pitch circle. From 
Mechanics of Materials it is known that the moment of flexure, 

PL= — ; in which S is the unit stress in the outer fibre ; /is the 
c 

bd' 6 
moment of inertia of the cross section, = — ; and c is the distance 

\L 

d XT ni Sbd* 

-. Hence, PI = —77-; 



from the neutral axis to the outer fibre, = -. Hence, PI 



6 PI 
S = j-r i . Assume a value for diametral pitch, find the correspond- 
ing values of b, d, and I from table on page 58, and substitute in the 
above equation. S now becomes known, and may be compared 
with the ultimate strength of the material of the gear, = S v If the 

factor of safety, = -^, is a proper value, the assumed pitch is right. 
If not, another pitch may be assumed and checked as before. 



58 



MACHINE DESIGN. 



Table I. — For Use in Designing Gears. 



Diametral 


Circular 


Thickness 
of Tooth 


Width 


Safe Stress 

for Cast 
Iron Gear 


Safe Unit 
Stress for 
Cast Iron 


Safe Stress 

for Cast 
Steel Gear 


Safe Unit 
Stress for 
Cast Steel 


Pitch 


Pitch 


on the 


of Face 


Gear 


Gears 






Pitch Line 
d 


b 


Factor of 
Safety^-io 


Factor of 
Safety-^io 


Factor of 
Safety=6 


Factor of 
Safety=6 


% 


6.283 


3.141 


20 


15250 


763 


61000 


3033 


% 


4.189 


2.094 


13 


6590 


507 


26390 


2030 


1 


3.141 


1.571 


9 


3442 


382 


13770 


1530 


m 


2.513 


1.256 


7K 


2250 


305 


9000 


1220 


ik 


2.094 


1.047 


6 


1530 


255 


6120 


1020 


m 


1 . 795 


.897 


5K 


1200 


218 


4800 


872 


2 


1.571 


.785 


4K 


, 862 


192 


3450 


767 


2M 


1.396 


.698 


4 


684 


170 


2738 


682 


2y 2 


1.256 


.628 


3^ 


532 


152 


2130 


610 


2% 


1.142 


.571 


3 


417 


139 


1668 


556 


3 


1.047 


.523 


2% 


318 


127 


1272 


509 


3K 


.897 


.449 


2 


216 


108 


864 


437 


4 


.785 


.393 


w 


168 


9^ 


672 


383 


5 


.628 


.314 


1% 


124 


, 76 


498 


306 


6 


.523 


.262 


IK 


96 


63 


384 


253 


7 


449 


.224 


1% 


74 


54 


296 


216 


8 


.393 


.196 


IK 


60 


48 


240 


192 


9 


.349 


.174 


IK 


48 


42 


191 


170 


10 


.314 


.157 


lhe 


41 


38 


163 


153 


12 


.262 


.131 


1«16 


30 


32 


120 


128 


14 


224 


.112 


13l6 


22 


27 


88 


109 



55. The work of approximation may be avoided by the use of 
Table I. Column 1 gives values of diametral pitch. Column 2 
gives values of circular pitch. Column 3 gives values of d. 
Column 4 gives values of width of face, = b, corresponding to 
good practice. If this value of b is accepted, the table may be 
used as follows : The maximum working force at the pitch 
surface of the gear is found as above. If this value is found in 
column 5, the value of diametral pitch horizontally opposite in 
column 1 may be used for a cast iron gear, with a factor of safety of 
10. If the value is not found in column 5, take the next greater 
value, and use the corresponding pitch. This will slightly increase 
the factor of safety. 

If the conditions of the design require some different value for b, 
divide the maximum working force at the pitch surface by the 




3 f^L^J. 




&_ r/a.3Z 



3 



TOOTHED WHEELS, OE GEAKS. 59 

width of face, find this value, or the next greater, in column 6, and 
use the corresponding value of pitch. 

If the pitch thus determined is too large for the design, a steel 
casting may be used, and the pitch will be determined by use of 
column 7 or 8 as above. 

56. Non-Circular Wheels. — Only circular centrodes or pitch 
curves correspond to a constant velocity ratio ; and by making the 
pitch curves of proper form, almost any variation in the velocity 
ratio may be produced. Thus a gear whose pitch curve is an ellipse, 
rotating about one of its foci, may engage with another elliptical 
gear, and if the driver has a constant angular velocity, the follower 
will have a constantly varying angular velocity. If the follower is 
rigidly attached to the crank of a slider crank chain, the slider will 
have a quick return motion. This is sometimes used for shapers 
and slotting machines. When more than one fluctuation of veloc- 
ity per revolution is required, it may be obtained by means of 
" lobed gears " ; i. e., gears in which the curvature of the pitch 
curve is several times reversed. If a describing circle be rolled on 
these non-circular pitch curves, the tooth outlines will vary in 
different parts ; hence, in order to cut such gears, many cutters 
would be required for each gear. Practically, this would be too 
expensive ; and when such gears are used the pattern is accurately 
made, and the cast gears are used without "tooling" the tooth 
surfaces. 

57. Bevel Gears. — All transverse sections of spur gears are the 
same, and their axes intersect at infinity. Spur gears serve to 
transmit motion between parallel shafts. It is necessary also to 
transmit motion between shafts whose axes intersect. In this case 
the pitch cylinders become pitch cones ; the teeth are formed upon 
these conical surfaces, the resulting gears being called bevel gears. 
To illustrate, let a and b, Fig. 57, be the axes between which the 
motion is to be transmitted with a given velocity ratio. This ratio 
is equal to the ratio of the length of the line A to that of B. Draw 
a line CD parallel to a, at a distance from it equal to the length of 



60 MACHINE DESIGN. 

the line A. Also draw the line CE parallel to b, at a distance from 
it equal to the length of the line B. Join the point of intersection 
of these lines to the point C, the intersection of the given axes. 
This locates the line CF, which is the line of contact of two pitch 
cones which will roll together to transmit the required velocity ratio. 

For — -= „, and if it be supposed that there are frusta of cones so 
nc B 

thin that they may be considered cylinders, their radii being equal 
to me and nc, it follows that they would roll together, if slipping be 
prevented, to transmit the required velocity ratio. But all pairs of 

7YIC 

radii of these pitch cones have the same ratio, = — ■, and therefore 

r nc 

any pair of frusta of the pitch cones may be used to roll together 
for the transmission of the required velocity ratio. To insure this 
result, slipping must be prevented, and hence teeth are formed 
upon the selected frusta of the pitch cones. The theoretical deter- 
mination of these may be explained as follows : 

1st. Cycloidal Teeth. — If a cone A (Fig. 58), be rolled upon 
another cone, B, an element be of the cone A will generate a conical 
surface, and a spherical section of this surface, adb, is called a 
spherical epicycloid. Also if a cone, A (Fig. 59), roll on the inside 
of another cone, 0, an element be of A will generate a conical sur- 
face, a spherical section of which, bda, is called a spherical hypo- 
cycloid. If now the three cones, B, C, and A, roll together, always 
tangent to each other on one line, as the cylinders were in the case 
of spur gears, there will be two conical surfaces generated by an 
element of A ; one upon the cone B, and another upon the cone C. 
These may be used for tooth surfaces to transmit the required con- 
stant velocity ratio. Because, since the line of contact of the cones 
is the axo * of the relative motion of the cones, it follows that a 
plane normal to the motion of the describing element of the gen- 
erating cone at any time, will pass through this axo. And also, 
since the describing element is always the line of contact between 

* An axo is an instantaneous axis, of which a centro is a projection. 



TOOTHED WHEELS, OR GEARS. 61 

the generated tooth surfaces, the normal plane to the line of contact 
of the tooth surfaces always passes through the axo, and the con- 
dition of rotation with a constant velocity ratio is fulfilled. 

2d. Involute Teeth. — If two pitch cones are in contact along 
an element, a plane may be passed through this element, making 
an angle (say 75°) with the plane of the axes of the cones. Tan- 
gent to this plane there may be two cones, whose axes coincide 
with the axes of the pitch cones. If a plane be supposed to wind 
off from one base cone upon the other, the line of tangency of the 
plane with one cone will leave the cone and advance in the plane 
toward the other cone, and will generate simultaneously upon the 
pitch cones conical surfaces, and spherical sections of these surfaces 
will be spherical involutes. These surfaces may be used for tooth 
surfaces, and will transmit the required constant velocity ratio, be- 
cause the tangent plane is the constant normal to the tooth sur- 
faces at their line of contact, and this plane passes through the axo 
of the pitch cones. 

To determine the tooth surfaces with perfect accuracy, it would 
be necessary to draw the required curves on a spherical surface, 
and then to join all points of these curves to the point of intersec- 
tion of the axes of the pitch cones. Practically this would be im- 
possible, and an approximate method is used. 

If the frusta of pitch cones be given, B and C, Fig. 60, then 
points in the base circles of the cones, as L, M, and K, will move 
always in the surface of a sphere whose projection is the circle 
LAKM. Properly, the tooth curves should be laid out on the sur- 
face of this sphere, and joined to the centre of the sphere to gen- 
erate the tooth surfaces. Draw cones LGM and MHK tangent to 
the sphere on circles represented in projection by lines LM and MK. 
If now tooth curves be drawn on these cones, with the base circle of 
the cones as pitch circles, they will very closely approximate the 
tooth curves that should be drawn on the spherical surface. But a 
cone may be cut along one of its elements and rolled out, or de- 
veloped, upon a plane. Let MDH be a part of the cone MHK, de- 
veloped, and let MNG be a part of the cone MGL, developed. The 



62 MACHINE DESIGN. 

circular arcs MD and MN may be used just as pitch circles are in 
the case of spur gears, and the teeth may be laid out in exactly the 
same way, the curves being either cycloidal or involute, as required. 
Then the developed cones may be wrapped back, and the curves 
drawn may serve as directrices for the tooth surfaces, all of whose 
elements converge to the centre of the sphere of motion. 

58. Theteeth of spur gears may be cut by means of milling cut- 
ters, because all transverse sections are alike ; but with bevel gears 
the conditions are different. The tooth surfaces are conical sur- 
faces, and therefore the curvature varies constantly from one end of 
the tooth to the other. Also the thickness of the tooth and the 
width of space vary constantly from one end to the other. But the 
curvature and thickness of a milling cutter cannot vary, and there- 
fore a milling cutter cannot cut an accurate bevel gear. Small 
bevel gears are, however, cut with milling cutters with sufficient 
accuracy for practical purposes. The cutter is made as thick as 
the narrowest part of the space between the teeth, and its curvature 
is made that of the middle of the tooth. Two cuts are made for 
each space. Let Fig. 61 represent a section of the cutter. For the 
first cut it is set relatively to the gear blank, so that the pitch point 
a of the cutter travels toward the apex of the pitch cone, and for 
the second cut so that the pitch point b travels toward the apex of 
the pitch cone. This method gives an approximation to the re- 
quired form. Gears cut in this manner usually need to be filed 
slightly before they work satisfactorily. Bevel gears with abso- 
lutely correct tooth surfaces may be made by planing. Suppose a 
planer in which the tool point travels always in some line through 
the apex of the pitch cone. Then suppose that as it is slowly fed 
down the tooth surface, it is guided along the required tooth curve 
by means of a templet. From what has preceded it will be clear 
that the tooth so formed will be correct. Planers embodying these 
principles have been designed and constructed by Mr. Corliss of 
Providence, and Mr. Gleason of Rochester, with the most satisfac- 
tory results. 

59. Design of Bevel Gears. — Given energy to be transmitted, rate 




r/*.J8 



/a \ 



K 



W 



N Ftf . 60. 



TOOTHED WHEELS, OR GEARS. 63 

of rotation of one shaft, velocity ratio, and angle between axes ; to 
design a pair of bevel gears. Locate the intersection of axes, 0, 
Fig. 62. Draw the axes OA and OB, making the required angle 
with each other. Locate OC, the line of tangency of the pitch 
cones, by the method given on page 59. Any pair of frusta of the 
pitch cones may be selected upon which to form the teeth. Special 
conditions of the problem usually dictate this selection approx- 
imately. Suppose that the inner limit of the teeth may be con- 
veniently at D. Then make DP, the width of face, = DO -v- 2. Or, 
if P is located by some limiting condition, lay off PD'*=P0 —- 3. 
In either case the limits of the teeth are defined tentatively. Now 
from the energy and the number of revolutions of one shaft (either 
shaft may be used), the moment of torsion may be found. The 
mean force *at the pitch surface == this torsional moment -s- the 
mean radius of the gear ; i. e., the radius of the point M, Fig. 62, 
midway between P and D. The pitch corresponding to this force 
may be found from Table I. This would be the mean pitch of the 
gears. But the pitch of bevel gears is measured at the large end, 
and diametral pitch varies inversely as the distance from 0. In 
this case the distances of M and P from are to each other as 5 is 
to 6. Hence the value of diametral pitch found from the table 

5 
X ■* — the diametral pitch of the gear. If this value does not cor- 
respond with any of the usual values of diametral pitch, the next 
smaller value may be used. This would result in a slightly in- 
creased factor of safety. If the diametral pitch thus found, multiplied 
by the diameter corresponding to the point P, does not give an integer 
for the number of teeth, the point P may be moved outward along 
the line OC, until the number of teeth becomes an integer. This 
also would result in slight increase of the factor of safety. The 
point Pis thus finally located, the corrected width of face = P0-^- 3, 
and the pitch is known. The drawing of the gears may be com- 
pleted as follows : Draw AB perpendicular to PO. With A and B 
as centres, draw the arcs PE and PF. Use these as pitch arcs, and 
draw the outlines of two or three teeth upon them, with cycloidal 



64 MACHINE DESIGN. 

or involute curves as required. These will serve to show the form 
of tooth outlines. From P each way along the line AB lay off the 
addendum and the clearance. From the four points thus located 
draw lines toward 0, terminating in the line DG. The tops of teeth 
and the bottoms of spaces are thus denned. Layoff upon AB below 
the bottoms of the spaces, a space about equal to the thickness of 
the tooth on the pitch circle. This gives a ring of metal to support 
the teeth. Join this to a properly proportioned hub as shown. 
The plan and elevation of each gear may now be drawn by the 
ordinary methods of projection. 

60. Skew Bevel Gears. — Spur gears serve to communicate motion 
between parallel axes, and bevel gears between axes that intersect. 
But it is sometimes necessary to communicate motion between axes 
that are neither parallel nor intersecting. If the parallel axes are 
turned out of parallelism, or if intersecting axes are moved into 
different planes, so that they no longer intersect, the pitch surfaces 
become hyperbolic paraboloids in contact with each other along a 
straight line, which is the generatrix of the pitch surfaces. These 
hyperbolic paraboloids rolled upon each other, circumferential 
slipping being prevented, will transmit motion with a constant 
velocity ratio. There is, however, necessarily a slipping of the 
elements of the surfaces upon each other parallel to themselves. 
Teeth may be formed on these pitch surfaces, and they may be 
used for the transmission of motion between shafts that are not 
parallel nor in the same plane. Such gears are called " Skew Bevel 
Gears." The difficulties of construction and the additional friction 
due to slipping along the elements, make them undesirable in prac- 
tice, and there is seldom a place where they cannot be replaced by 
some other form of connection. 

A very complete discussion of the subject of Skew Bevel Gears 
may be found in Prof. McCord's " Kinematics." 

61. Spiral Gearing. — If line contact is not essential there is much 
wider range of choice of gears to connect shafts which are neither 
parallel nor intersecting. A and B, Fig. 63, are axes of rotation in 
different planes, both planes being parallel to the paper. Let EF 



TOOTHED WHEELS, OR GEARS. 65 

and GH be cylinders on these axes, tangent to each other at the 
point S. Any line may now be drawn through S either between A 
and B, or coinciding with either of them. This line, say DS, may 
be taken as the common tangent to helical or screw lines drawn on 
the cylinders EF and GH ; or helical surfaces may be formed on 
both cylinders, DS being their common tangent at S. Spiral Gears 
are thus produced. Each one is a portion of a many-threaded 
screw. The contact in these gears is point contact ; in practice the 
point of contact becomes a very limited area. 

62. When the angle between the shafts is made equal to 90°, 
and one gear has only one or two threads, it becomes a special case 
of spiral gearing known as Worm Gearing. In this special case the 
gear with a single or double thread is called the worm, while the 
other gear, which is still a many-threaded screw, is called the worm, 
wheel. If a section of a worm and worm wheel be made on a plane 
passing through the axis of the worm, and normal to the axis of 
the worm wheel, the form of the teeth will be the same as that of a 
rack and pinion ; in fact the worm, if moved parallel to its axis, 
would transmit rotary motion to the worm wheel. From the con- 
sideration of racks and pinions it follows that if the involute sys- 
tem is used, the sides of the worm teeth will be straight lines. 
This simplifies the cutting of the worm, because a tool may be used 
capable of being sharpened without special methods. If the worm 
wheel were only a thin plate the teeth would be formed like those 
of a spur gear, of the same pitch and diameter. But since the 
worm wheel must have greater thickness, and since all other sec- 
tions parallel to that through the axis of the worm, as CD and AB, 
Fig. 64, show a different form and location of tooth, it is necessary 
to make the teeth of the worm wheel different from those of a spur 
gear, if there is to be contact between the worm and worm wheel 
anywhere except in the plane EF, Fig. 64. This would seem to in- 
volve great difficulty, but it is accomplished in practice as follows : 
A duplicate of the worm is made of tool steel, and " flutes" are cut 
in it parallel to the axis, thus making it into a cutter, which is tem- 
pered. It is then mounted in a frame in the same relation to the 



66 MACHINE DESIGN. 

worm wheel that the worm is to have when they are linished and 
in position for working. The distance between centres, however, 
is somewhat greater, and is capable of being gradually reduced. 
Both are then rotated with the required velocity ratio by means of 
gearing properly arranged, and the cutter or "hob" is fed against 
the worm wheel till the distance between centres becomes the 
required value. The teeth of the worm wheel are " roughed out " 
before they are "hobbed/' By the above method the worm is made 
to cut its own worm wheel.* 

Fig. 65 represents the half section of a worm. If it is a single 
worm the thread A, in going once around, comes to B ; twice 
around, to C ; and so on. If it is a double worm the thread A, in 
going once around, comes to C, while there is an intermediate 
thread, B. It follows that if the single worm turns through one 
revolution it will push one tooth of the worm wheel with which it 
engages, past the line of centres ; while the double worm will push 
two teeth of the worm wheel past the line of centres. The single 
worm, therefore, must make as many revolutions as there are teeth 
in the worm wheel, in order to cause one revolution of the worm 
wheel ; while for the same result the double worm only needs to 
make half as many revolutions. The ratio of the angular velocity 
of a single worm to that of the worm wheel with which it engages 

is = -, in which n equals the number of teeth in the worm wheel. 

For the double worm this ratio is -. 

A 

Worm gearing is particularly well adapted for use where it is 
necessary to get a high velocity ratio in limited space. 

The pitch of a worm is measured parallel to the axis of rotation. 
The pitch of a single worm is p, Fig. 65. It is equal to the circular 
pitch of the worm wheel. The pitch of a double worm is p l = 2p=^ 
2 X circular pitch of the worm wheel. 

* This subject is fully treated in Unwin's " Elements of Machine Design," 
and in Brown & Sharpe's " Treatise on Gearing." 



TOOTHED WHEELS, OR GEARS. 67 

63. Design of Worm Gears. — Let E = energy to be transmitted 
through the worm wheel per minute ; N = number of revolutions 
per minute; R = pitch radius of the worm wheel; F=- force at 
pitch surface of the worm wheel. Then 

E = 2-RN X F = space factor X force factor, 
_ _. E torsional moment 

and F= ^RN= R ■ 

Hence, if E, R, and N are given, F becomes known, and an approx- 
imate value for pitch may be found from Table I. If p is the 

diametral pitch thus found, the corresponding circular pitch — - 

= pitch of a single worm to mesh with the worm wheel.* The 

pitch of a double worm to mesh with the worm wheel = — . The 

p 

number of teeth, n, in the worm wheel = 2Rp. For a single worm 

71 7\ 

the velocity ratio = - ; for double worm == -. The rate of rotation 

Nn 
of the single worm = Nn ; of the double worm = — . From the 

energy, E v to be transmitted per minute by the worm shaft, and the 

E 

rate of rotation, N, the moment of torsion is found =—~^. From 
' " 2^ 

this a suitable belt driving mechanism may be designed by methods 

to be given later. 

64. When the worm and worm wheel are determined, a working 
drawing may be made as follows : Draw AB, Fig. 66, the axis of the 
worm wheel, and locate 0, the projection of the axis of the worm, 
and P, the pitch point. With as a centre draw the pitch, full 
depth, and addendum circles, G, H, and K ; also the arcs CD and 
EF, bounding the tops of the teeth and the bottoms of the spaces of 
the worm wheel. Make the angle p = 90°. Below EF lay off a 
proper thickness of metal to support the teeth, and join this by the 

* This value must be made such that it may be cut in an ordinary lathe. 
See next page. 



d 



68 MACHINE DESIGN. 

web LM to the hub N. The tooth outlines in the other sectional 
view are drawn exactly as for an involute rack and pinion. Full 
views might le drawn, but they involve difficulties of construction, 
and do not give any additional information to the workman. 

65. Solution from Other Data. — Two shafts, at right angles to 
each other in different planes, are to be connected by means of 
worm gearing. The maximum distance between them is fixed, and 
its value, d, is given. The required velocity ratio, r, the rate of 
rotation of the worm shaft, N, and the energy to be transmitted per 
minute, E, are also given. It is required to design the gears. A 
single thread worm is to be used. Let E = 33000 ft. lbs. per min- 
ute ; r — 40, i. e., the worm makes 40 revolutions per revolution of 
the worm wheel ; d = 8"; N =280 revolutions per minute. The 
velocity ratio depends upon the pitch of the worm, but not upon its 
diameter. Because, whatever the diameter of the worm, it pushes 
one tooth of the worm wheel past the line of centres (QR, Fig. 66) 
each revolution. The pitch diameter of the worm may therefore be 
any convenient value. The worm wheel must have 40 teeth in 
order that the single thread worm shall turn 40 times to turn it 
once. The number of threads per inch of the worm, measured 
axially, is the reciprocal of the pitch. This should be such a value 
that it may be cut in an ordinary lathe without special arrange- 
ment of the change gears. Lathes of medium size are capable of 
cutting 1, 2, 3, 4, etc., threads per inch. The circular pitch may 
therefore be 1, 0*5, 0'334, 0*25, etc. In the case considered suppose 
that the pitch diameter of the worm may conveniently be about 
3S". The corresponding pitch radius of the worm wheel = 8"— 1*5" 
= 6'5". Since there must be 40 teeth, it follows that the circular 

pitch, = --77T~ = 1'021. This should be exactly 1, as indicated 

above. Let circular pitch = 1, and check for strength. The tooth 
may be considered as a cantilever, as in the case of spur gear teeth. 

Then S = r-^, in which P = force at the pitch surface, I = depth 

of space, b = distance corresponding to the arc EF, Fig. 66, and 



TOOTHED WHEELS, OR GEARS. 69 

d = thickness of tooth at pitch circle = \ circular pitch. — To find 

P. The energy = 33,000 ft.lbs. per minute ; N = rate of rota- 

280 
tion of the worm wheel shaft = — — = 7 ; the pitch radius of the 

40 r 

40 
worm wheel corrected for 1 pitch =•= — - = 6*37" = 0*53 feet ; the 

torsional moment at the worm wheel shaft = _ _. = Pr. Hence, 

2*^ 

D 33000 33000 1 .. . .. _ _.„ . n _ 

P= -^ = ^X0^X"7^ 14H lb8 - ; Z = °' 64; d = °' 6 ' 

To find b. The pitch radius of the worm corrected for 1 pitch = 

8 — 6*37 = 1*63. The radius of the outside of the worm = 1*63 + 

addendum, 0*318, = 1*948 ; the arc subtended by 90° on this radius 

fi v 141 4 v 0-64 
= 90* X 00174 X 1*948 = 3*05 =- b. Hence S = * .. *- = 

3*0o X 0*25 

7121 lbs. If cast iron were to be used, this would give a factor of 

safety = = 2*8. This is too small, and a larger circular 

pitch would need to be used, and the worm would have to be cut in 
a lathe capable of cutting less than 1 thread per inch. If steel 

casting is an allowable material the factor of safety would = „,^., 

J /121 

= 7 -f- . This is a proper value, and the design is correct. 

This method of design applies when the worm wheel is cut by 
a "hob." 

66. Compound Spur Gear Chains. — Spur gear chains may be com- 
pound, i. e.j they may contain links which carry more than two 
elements. Thus in Fig. 67 the links a and d each carry three 
elements. In the latter case the teeth of d must be counted as two 
elements, because by means of them d is paired with both b and c. 
In the case of the three-link spur gear chain the wheels b and c 
meshed with each other, and a point in the pitch circle of c moved 
with the same linear velocity as a point in the pitch circle of b, but 
in the opposite direction. In Fig. 67 points in all the pitch circles 
have the same linear velocity, since the motion is equivalent to 



70 MACHINE DESIGN. 

rolling together of the pitch circles without slipping ; but c and ?> 
now rotate in the same direction. Hence it is seen that the intro- 
duction of the wheel d has reversed the direction of rotation, with- 
out changing the velocity ratio. The size of the wheel d, which is 
called an "idler,"' has no effect upon the motion of c and b. It 
simply receives, upon its pitch circle, a certain linear velocity from 
c, and transmits it unchanged to b. Hence the insertion of any 
number of idlers does not affect the velocity ratio of c to b, but each 
added idler reverses the direction of the motion. Thus, with an odd 
number of idlers, c and b will rotate in the same direction ; and 
with an even number of idlers, c and b will rotate in opposite direc- 
tions. 

If parallel lines be drawn through the centres of rotation of a 
pair of gears, and if from the centres distances be laid off on these 
lines inversely proportional to the angular velocities of the gears, 
then a line joining the points so determined will cut the line of 
centres in a point which is the centro of the gears. . In Fig. 67, 
since the rotation is in the same direction, the lines have to be laid 
off on the same side of the line of centres. The pitch radii are 
inversely proportional to the angular velocities of the gears, and 
hence it is only necessary to draw a tangent to the pitch circles of b 
and e, and the intersection of this line with the line of centres is the 
centro, be, of c and b. The centroids of c and b are c 1 and b x , circles 
through the point be, about the centres of c and b. Obviously, this 
four link mechanism may be replaced by a three link mechanism, 
in which c is an annular wheel meshing with a pinion b. The four 
link mechanism is more compact, however, and usually more con- 
venient in practice. 

67. The other principal form of spur gear chain is shown in 
Fig. 68. The wheel d has two sets of teeth of different pitch diam- 
eter, one pairing with e, and the other with b. The point bd now 
has a different linear velocity from cd, greater or less in proportion 
to the ratio of the radii of those points. The angular velocity ratio 
may be obtained as follows : 



TOOTHED WHEELS, OR GEARS. 71 

angular veloc. d Ccd 



also 



Multiplying, 



angular 


veloc. 


c 


Dcd' 


angular 


veloc. 


b 


Dbd 


angular 


veloc. 


d 


~ Bbd' 


angular veloc. b 


_ Ccd X Dbd 



angular veloc. c Dcd X Bbd ' 



The numerator of the last term consists of the product of the radii 
of the " followers " ; and the denominator consists of the product of 
the radii of the " drivers." The diameters or numbers of teeth 
could be substituted for the radii. 

In general, the angular velocity of the first driver is to the 
angular velocity of the last follower as the product of the number 
of teeth of the followers is to the product of the number of teeth of 
the drivers. This applies equally well to compound spur gear 
trains that have more than three axes. Therefore, in any spur gear 
chain the velocity ratio equals the product of the number of teeth 
in the followers divided by the product of the number of teeth in 
the drivers. The direction of rotation is reversed if the number of 
intermediate axes is even, and is not reversed if the number is odd. 
If the train includes annular gears their axes would be omitted 
from the number, because annular gears do not reverse the direc- 
tion of rotation. 



CHAPTER VI. 



CAMS. 



68. A machine part of irregular outline, as A, Fig. 69, may 
rotate or vibrate about an axis 0, and communicate motion by line 
contact to another machine part, B. A is called a cam. A cylin- 
der A, Fig. 70, having a groove of any form in its surface, may 
rotate about its axis, CD, and communicate motion to another 
machine part, B. A is a cam. A disc A, Fig. 71, having a groove 
in its face, may rotate about its axis, 0, and communicate motion 
to another machine part, B. A is a cam. In fact it is only a mod- 
ification of A, Fig. 69. In designing cams it is customary to con- 
sider a number of simultaneous positions of the driver and follower. 
The cam curve can usually be drawn from data thus obtained. 

69. Case I. — The follower is guided in a straight line, and the 
contact of the cam with the follower is always in this line. The line 
may be in any position relatively to the centre of rotation of the 
cam ; hence it is a general case. The point of the follower which 
bears on the cam is constrained to move in the line MN, Fig. 72. 
is the centre of rotation of the cam. About 0, as a centre, draw 
a circle tangent to MN at J. Then A, B, C, etc., are points in the 
cam. When the point A is at J the point of the follower which 
bears on the cam must be at A' ; when B is at / the follower point 
must be at B' ; and so on through an entire revolution. Through 
A, B, C, etc., draw lines tangent to the circle. With as a centre, 
and OA' as a radius, draw a circular arc A' A", intersecting the tan- 
gent through A at A". Then A" will be a point in the cam curve. 
For, if A returns to J, AA" will coincide with JA', A" will coincide 
with A', and the cam will hold the follower in the required position. 



CAMS. 73 

The same process for the other positions locates other points of the 
cam curve. A smooth curve drawn through these points is the 
required cam outline. Often, to reduce friction, a roller attached 
to the follower rests on the cam, motion being communicated 
through it. The curve found as above will be the path of the axis 
of the roller. The cam outline will then be a curve drawn inside 
of, and parallel to, the path of the axis of the roller, at a distance 
from it equal to the roller's radius. Contact between the follower 
and the cam is not confined to the line MN if a roller is used. 

70. Case II — The cam engages with a surface of the follower, 
and this surface is guided so that all of its positions are parallel. 
The method given is due to Professor J. H. Barr. 0, Fig 73, is the 
centre of rotation of the cam. The follower surface occupies the 
successive positions 1, 2, 3, etc., when the lines A, B, C, etc. of the 
cam coincide with the vertical line through G. It is required to 
draw the outline of a cam to produce the motion required. Pro- 
duce the vertical line through 0, cutting the positions of the 
follower surface in A', B', C, etc. With as a centre and radii 
OB', OC, etc., draw arcs cutting the lines B, C, D, etc. in the points 
B", C" , D" , etc. Position 1 is the lowest position of the follower 
surface ; therefore A must be in contact with the follower surface in 
the vertical line through 0, because if the tangency be at any other 
point the motion in one direction or the other will lower the fol- 
lower, which is not allowable. A is therefore one point in the cam 
curve. Draw a line MN through B" at right angles to B"0, and 
rotate B"0 till it coincides with B'O. Then the line MN will 
coincide with the position of the follower surface 2B'. But the cam 
curve must be tangent to this line when B coincides with B'O, and 
therefore the line MN is a line to which the cam curve must be tan- 
gent. Similar lines may be drawn through the points C", D" , etc. 
Each will be a line to which the cam curve must be tangent. There- 
fore, if a smooth curve be drawn tangent to all these lines, it will 
be the required cam outline. 

71. Case III — This is the same as Case II, except that the posi- 
tions of the follower surface instead of being parallel, converge to a 
point, 0, Fig. 74, about which the follower vibrates. The solution 



74 MACHINE DESIGN. 

is the same as in Fig. 73, except that the angle between the lines 
corresponding to MN, Fig. 73, and the radial lines, instead of being 
a right angle, equals the angle between the corresponding position 
of the follower surface and the vertical. 

In these cases the cam drives the follower in only one direction ; 
the force of gravity, the expansive force of a spring, or some other 
force must hold it in contact with the cam. To drive the follower 
in both directions, the cam surface must be double, i. e., it takes the 
form of a groove engaging with a pin or roller attached to the fol- 
lower, as in Fig. 71. The foregoing principles apply to the laying 
out of the curves. 

72. Case IV. — To lay out a cam groove on the surface of a cylin- 
der. — -A, Fig. 75, is a cylinder which is to rotate continuously about 
its axis. B can only move parallel to the axis of A. B may have a 
projecting roller to engage with a groove in the surface of A. CD 
is the axis of the roller in its mid-position. EF is the development 
of the surface of the cylinder. During the first quarter revolution 
of A, CD is required to move one inch toward the right with a 
constant velocity. Lay off GH = 1", and HJ = \KF, locating /. 
Draw GJ, which will be the middle line of the cam groove. During 
the next half revolution of A, the roller is required to move two 
inches toward the left with a uniformly accelerated velocity. Lay 
off JL = 2", and LM=^KF. Divide LM into any number of 
equal parts, say 4. Divide JL into 4 parts, so that each is greater 
than the preceding one by an equal increment. This may be done 
as follows : l-j-2-}-3-r-4=10. Lay off from J, 01 JL, locating a; 
then 0*2 JL from a, locating b ; and so on. Through a, b, and c 
draw vertical lines ; through m, n, and o draw horizontal lines. 
The intersections locate d, e, and /. Through these points draw 
the curve from J to M, which will be the required middle line of 
the cam groove. During the remaining quarter revolution the 
roller is required to return to its starting point with a uniformly 
accelerated velocity. The curve MN is drawn in the same way as 
JM. On each side of the line GJMN lay off parallel lines, their 
distance apart being equal to the diameter of the roller. Wrap EF 
upon the cylinder, and the required cam groove is located. 



CHAPTER VII. 

BELTS. 

73. Transmission of Motion by Belts. — In Fig. 76, let A and B be 

two cylindrical surfaces, free to rotate about their axes ; let CD be 
their common tangent, and let CD represent an inextensible con- 
nection between the two cylinders. Since it is inextensible, the 
points D and 0, and hence the surfaces of the cylinders, must have 
the same linear velocity. Two points having the same linear 
velocity, and different radii, have angular velocities which are in- 
versely proportional to their radii. Hence, since the surfaces of 
the cylinders have the same linear velocity, their angular velocities 
are inversely proportional to their radii. This is true of all cylinders 
connected by inextensible connectors. Suppose the cylinders to 
become pulleys, and the tangent line to become a belt. Let CD' 
be drawn ; this becomes a part of the belt, making it endless, and 
rotation may be continuous. The belt will remain always tangent 
to the pulleys, and will transmit such rotation that the angular 
velocity ratio will constantly be the inverse ratio of the radii of the 
pulleys. 

The case considered corresponds to a crossed belt, but the same 
reasoning applies to an open belt. See Fig. 77. A and B are two 
pulleys, and CDD'C'C is an open belt. Since the points C and D 
are connected by a belt that is practically inextensible, the linear 
velocity of C and D is the same ; therefore the angular velocities of 
the pulleys are to each other inversely as their radii. If the pulleys 
in either case were pitch cylinders of gears the conditions of velocity 
would be the same. In the first case, however, the direction of 
motion is reversed, while in the second case it is not. Hence the 



76 MACHINE DESIGN. 

first corresponds to gears meshing directly with each other, while 
the second corresponds to the case of gears connected by an idler, 
or to the case of an annular gear and pinion. 

Of course it is necessary that a belt should have some thickness ; 
and, since the centre of pull is the centre of the belt, it is necessary 
to add to the radius of the pulley half of the thickness of the belt. 
The motion communicated by means of belting, however, does not 
need to be absolutely correct, and therefore in practice it is usually 
customary to neglect the thickness of the belt. The proportioning 
of pulleys for the transmission of any required velocity ratio is now 
a very simple matter. 

Illustration. — A line shaft runs 150 revolutions per minute, 
and is supported by hangers with 16" "drop." It is required to 
transmit motion from this shaft to a dynamo to run 1800 revolu- 
tions per minute. A 30" pulley is the largest that can be con- 
veniently used with 16" hangers. Let x = the diameter of required 
pulley for the dynamo ; then from what has preceded x -r- 30 = 
150 -5- 1800, and therefore x == 2*5". But a pulley less than 4" di- 
ameter should not be used on a dynamo. Suppose in this case that 
it is 6". It is then impossible to obtain the required velocity ratio 
with one change of speed, i. c, with one belt. Two changes of 
speed may be obtained by the introduction of a counter shaft. By 
this means the velocity ratio is divided into two factors. If it is 
wished to have the same change of speed from the line shaft to the 
counter as from the counter to the dynamo, then each velocity ratio 
would be !/(1800 -s- 150) = j/12 = 3*46. But this gives an incon- 
venient fraction, and the factors do not need to be equal. Let the 
factors be 3 and 4. See Fig. 78. A represents the line shaft, B the 
counter, and C the dynamo shaft. The pulley on the line shaft is 
30", and the speed is to be three times as great at the counter, and 
therefore the pulley must have a diameter one-third as great, = 10", 
The pulley on the dynamo is 6" diameter and the counter shaft is 
to run one-fourth as fast, and therefore the pulley on the counter 
opposite the dynamo pulley must be four times as large as the 
dynamo pulley, = 24". 



BELTS. 77 

74. A belt may be shifted from one part of a pulley to another 
by means of pressure against the side which advances toward 
the pulley. Thus, if in Fig. 79 the rotation be as indicated by the 
arrow, and side pressure be applied at A, the belt will be pushed to 
one side, as is shown, and will consequently be carried into some 
new position on a pulley further to the left as it advances. Hence, 
in order that a belt may maintain its position on a pulley, the centre 
line of the advancing side of the belt must be perpendicular to the axis 
of rotation. When this condition is fulfilled the belt will run and 
transmit the required motion, regardless of the relative position of 
the shafts. 

75. In Fig. 80, the axes AB and CD are parallel to each other, 
the above stated condition is fulfilled, and the belt will run cor- 
rectly ; but if the axis CD were turned into some new position, as 
CD' , the side of the belt that advances toward the pulley E, cannot 
have its centre line in a plane perpendicular to the axis, and there- 
fore it will run off. But if a plane be passed through the line CD, 
perpendicular to the plane of the paper, then the axis may be swung 
in this plane in such a way that the necessary condition shall be 
fulfilled, and the belt will run properly. This gives what is known 
as a "twist" belt, and when the angle between the shafts becomes 
90°, it is a "quarter twist" belt. To make this clearer, see Fig. 81. 
Rotation is transmitted from A to B by an open belt, and it is 
required to turn the axis of B out of parallelism with that of A. The 
direction of rotation is as indicated by the arrows. Draw the line 
CD. If now the line CD be supposed to pass through the centre of 
the belt at C and D, it may become an axis, and the pulley B and 
the part of the belt FC may be turned about it, while the pulley A 
and the part of the belt ED remain stationary. During this motion 
the centre line of the part of the belt CF, which is the part that 
advances toward the pulley B when rotation occurs, is always in a 
plane perpendicular to the axis of the pulley B. The part ED, 
since it has not been moved, has also its centre line in a plane per- 
pendicular to the axis of A. Therefore, the pulley B may be swung 
into any angular position about CD as an axis, and the condition 
of proper belt transmission will not be interfered with. 



78 MACHINE DESIGN. 

76. If the axes intersect, the motion can be transmitted between 
them by belting only by the use of "guide" or "idler" pulleys. 
Let AB and CD, Fig. 82, be intersecting axes, and let it be required 
to transmit motion from one to the other by means of a belt run- 
ning on the pulleys E and F. Draw centre lines EK and FH 
through the pulleys. Draw the circle, G, of any convenient size, 
tangent to the lines EK and FH. In the axis of the circle 6r, let a 
shaft be placed on which are two pulleys, their diameters being 
equal to that of the circle G. These will serve as guide pulleys for 
the upper and lower sides of the belt, and by means of them the 
centre lines of the advancing parts of both sides of the belt will be 
kept in planes perpendicular to the axis of the pulley toward which 
they are advancing, the belts will run properly, and the motion 
will be transmitted as required. 

77. An analogy will be noticed between gearing and belting for 
the transmission of rotary motion. Spur gearing corresponds to an 
open or crossed belt, transmitting motion between parallel shaftsr 
Bevel gears correspond to a belt running on guide pulleys, trans- 
mitting motion between intersecting shafts. Skew bevel and spiral 
gears correspond to a " twist " belt, transmitting motion between 
shafts that are neither parallel nor intersecting. 

78. If a flat belt be put on a " crowning " pulley, as in Fig. 83, 
the tension on AB will be greater than on CD, the belt will tend to 
be shifted into the position shown by the dotted lines i£and F, and as 
rotation goes on, the belt will be carried toward the high part of the 
pulley, i. e.j it will tend to run in the middle of the pulley. This is 
the reason why nearly all belt pulleys, except those on which the 
belt has to be shifted into different positons, are turned "crown- 
ing." 

79. Cone Pulleys. — In performing different operations on a ma- 
chine or the same operations on materials of different degrees of 
hardness, different speeds are required. The simplest way of 
obtaining them is by the use of cone pulleys. One pulley has a 
series of steps, and the opposing pulley has a corresponding series 
of steps. By shifting the belt from one pair to another the velocity 



BELTS. 79 

ratio is changed. Since the same belt is used on all the pairs of 
steps, they must be so proportioned that the belt length for all the 
pairs shall be the same ; otherwise the belt would be too tight on 
some of the steps and too loose on others. Let the case of a crossed 
belt be first considered. The length of a crossed belt may be 
expressed by the following formula : Let L = length of the belt ; 
d == distance between centres of rotation ; R == radius of the larger 
pulley ; r = radius of the smaller pulley. See Fig. 84. Then L = 
2 } /d z —(R + r) 2 + (R + r)(* -f 2 arc whose sine is R + r -s- d). 
In the case of a crossed belt, if the size of steps be changed so that 
the sum of their radii remains constant, the belt length will be con- 
stant. For in the formula the only variables are R and V, and these 
terms only appear in the formula as R -f r ; but R -f- r is by 
hyothesis constant. Therefore any change that is made in the 
variables R and r, so long as their sum is constant, will not affect 
the value of the equation, and hence the belt length will be con- 
stant. It will now be easy to design cone pulleys for crossed belt. 
Suppose a pair of steps given to transmit a certain velocity ratio. 
It is required to find a pair of steps that will transmit some other 
velocity ratio, the length of belt being the same in both cases. Let 
r and r = radii of the given steps ; R and R' = radii of the required 
steps ; r -(- r = R + R' = a ; ; the velocity ratio of R to R' = b. 
There are two equations between R and R, R -i- R'= b, and R-\- R 
= a. Combining and solving, it is found that R' = a -=- ( 1 -j- />), 
and R = a — R. For an open belt the formula for length is : L = 
2 } /d' — (R — r) 2 + -(R +r)+2 (R — r) (arc whose sine is R — 
r-i-d). If R and r be changed as before, the term R — r would of 
course not be constant, and two of the terms of the equation would 
vary in value ; therefore the length of the belt would vary. The 
determination of cone steps for open belts therefore becomes a more 
difficult matter, and approximate methods are almost invariably 
used. 

80. The following graphical approximate method is due to Mr. 
C. A. Smith, and is given, with full discussion of the subject, in 
"Transactions of the American Society of Mechanical Engineers," 



80 MACHINE DESIGN. 

Vol. X, p. 269. Suppose first that the diameters of a pair of cone 
steps that transmit a certain velocity ratio are given, and that the 
diameters of another pair that shall serve to transmit some other 
velocity ratio are required. The distance between centres of axes is 
given. See Fig. 85. Locate the pulley centres and 0' , at the given 
distance apart ; about these centres draw circles whose diameters 
equal the diameters of the given pair of steps ; draw a straight line 
GH, tangent to these circles ; at J, the middle point of the line of 
centres, erect a perpendicular, and lay off a distance JK equal to 
the distance between centres, 0, multiplied by the experimentally 
determined constant 0314 ; about the point K so determined, draw a 
circular arc AB, tangent to the line GH. Any line drawn tangent to 
this arc will be the common tangent to a pair of cone steps giving the 
same belt length as that of the given pair. For example, suppose 
that OD is the radius of one step of the required pair ; about 0, 
with a radius equal to OD, draw a circle ; tangent to this circle and 
to the arc AB, draw a straight line DE ; about O and tangent to 
DE, draw a circle ; its diameter will equal that of the required step. 
But suppose that instead of having one step of the required pair 
given, to find the other corresponding as above, a pair of steps are 
required that shall transmit a certain velocity ratio, = r, with the 
same length of belt as the given pair. Suppose OD and O'E to rep- 
resent the unknown steps. The given velocity ratio equals r. But 
from similar triangles C€) -f- O'E = FO -v- FO' . Therefore r = 

FO C 4- x 

^yr,-, but FO — C 4- x, and FO' = x. Therefore r = , and 

F0 } ' x ' ■ 



x = — . Hence, with r and C given, the distance x may be 

found, such that if from F a line be drawm tangent to AB, the cone 
steps drawn tangent to it will give the velocity ratio, r, and a belt 
length equal to that of any pair of cones determined by a tangent 
to AB. The point F often falls at an inconvenient distance. The 
radii of the required steps may then be found as follows : Place a 
straight-edge tangent to the arc AB and measure the perpendicular 
distances from it to and 0'. The straight-edge may be shifted 
until these distances bear the required relation to each other. 



BELTS. 81 

81. Design of Belts. — Fig. 86 represents two pulleys connected 
by a belt. When no moment is applied tending to produce rota- 
tion this tension in the two sides of the belt is equal. Let T s repre- 
sent this tension. If now an increasing moment, represented by i??, 
be applied to the driver, its effect is to increase the tension in the 
lower side of the belt and to decrease the tension in the upper side. 
With the increase of Rl this difference of tension increases till it is 
equal to P, the force with which rotation is resisted at the surface of 
the pulley. Then rotation begins,* and continues as long as this 
equality continues; i. e., as long as 7\ — T. Z =P, in which 2\ = 
tension in the driving side, and T t = tension in the slack side. 
The tension in the driving side is increased at the expense of that 

T + T. 
in the slack side. Therefore — !— - — - — T. 6 . 

Ji 

T 
To find the value of ~. The increase in tension from the slack 

side to the driving side is possible because of the frictional resist- 
ance between the belt and pulley surface. Consider any element of 
the belt, ds, Fig. 88 (a). It is in equilibrium under the action of 
the following forces : T, the value of the varying tension corres- 
ponding to the section, acts upon one end of ds and is aided by dF. 
The force 7*+ (J Tacts upon the other end. From the action of 
these forces there results a normal pressure between ds and pulley, 
= pds, in which p = the pressure per linear unit of belt. Draw 
the force triangle, (b) Fig. 88. It is an isosceles triangle, and 

hence pds = (T + dT)0 ; but = -; .-. pds = ( T + dT ^ ; dT 

vanishes; :. V = 1. 
r 

Since the force triangle is an isosceles triangle, it follows that 

T -f d.T= T -\- dF ) hence dT— dF. Suppose that rotation occurs 

and that the belt slips upon the pulley at a rate corresponding to a 

* While the moving parts are being brought up t o speed the difference of 
tension must equal P-f- force necessary to produce the acceleration. 



82 MACHINE DESIGN. 

coefficient of friction/. Then dF=fpds, and since p 

/; dT=z f^±. but d8 = rd0, 
r 

.'. dT=fTdO; 



f3=>r.« 



T 

loge T. = f a; 

T 

-± = ef a , where e = the Naperian base. 

log I = 0-4343/a. 
a is in 7r measure and equals a in degrees X 0*0174. 
82. The following equations are established : 

r,-r 2 = p (i) 

T,+ T, = 2T 3 (2) 

I = e/« or log J> = 0-4843 fa (3) 

The right hand members of (1) and (3) can usually be deter- 
mined ; hence the value of T x (the maximum stress in the belt) 
may be found, and proper proportions may be given to the belt. If 
W foot-pounds per minute are to be transmitted, and the velocity 
of the rim of the pulley transmitting this power in feet per minute 
equal S, then the force P equals the work divided by the velocity; or, 

W 
P =. — . The value of « is found from the diameters of the pulleys 

and their distance between centres, and may usually be estimated 
accurately enough. The value of/, the coefficient of friction, varies 
with the kind of belting, the material and character of surface of 
the pulley, and with the rate of slip of the belt on the pulley. Ex- 
periments made at the laboratory of the Massachusetts Institute of 



BELTS. 83 

Technology, under the direction of Professor Lanza, indicate that 

for leather belting running on turned cast iron pulleys, the rate of 

slip for efficient driving is about three to four feet per minute ; and 

also that the coefficient of friction corresponding to this rate of slip 

is about 0-27. The value 03 may be used. If this value of / be 

used the slip will be kept within the above limits if the belt be put 

T 4- T. 
on with a proper initial tension, = T s = — — - — - , and the driving of 

the belt so designed will be satisfactory. 

83. Problem. — A single-acting pump has a plunger 8" = 0*666 
feet in diameter, whose stroke has a constant length of 10" = 0*833 
feet. The number of strokes per minute is 50. The plunger is 
actuated by a crank, and the crank shaft is connected by spur gears 
to a pulley shaft, the ratio of gears being such that the pulley shaft 
runs 300 revolutions per minute. The pulley which receives 
the power from the line shaft is 18" in diameter. The pressure in 
the delivery pipe is 100 lbs. per square inch. The line shaft runs 
150 revolutions per minute, and its axis is at a distance of 12 
feet from the axis of the pulley shaft. Since the line shaft runs 
half as fast as the pulley shaft, the diameter of the pulley on the 
line shaft must be twice as great as that on the pulley shaft, or 36". 
The work to be done per minute, neglecting the friction in the ma- 
chine, is equal to the number of pounds of water pumped per 
minute multiplied by the head in feet against which it is pumped. 
The number of cubic feet of water per minute equals the displace- 
ment of the plunger in cubic feet multiplied by the number of 

strokes per minute = — — -r X 0*833 X 50 = 14*5, and therefore 

the number of pounds of water pumped per minute = 14*5 X 62*4 
== 907. One foot vertical height or " head " of water corresponds 
to a pressure of 0*435 lbs. per square inch, and therefore 100 lbs. per 
square inch corresponds to a "head" of 100 -*- 0*435 = 230 feet. 
The work done per minute in pumping the water therefore is equal to 
907 lbs. X 230 feet = 208,610 foot-pounds. The velocity of the rim 
of the belt pulley — 300 X 1'5tt = 1410 feet per minute. Therefore 



84 MACHINE DESIGN. 

the force P=T 1 —T i = 208,610 ft.lbs. per minute -s- 1410 feet per 
minute = 147 lbs. .^ 

To rind «, see Fig. 89. T&0 = ?^ = ^ = 0'0625. There- 
fore ■-= 3° 35'; a = 180° — 20 = 180° — 7° 10' = 173° nearly ; a in 
7T measure == 178 X 0-0174 == 3*01. 

log ~ l = 0-4343 X/ X a == 0-4343 X 0-3 X 3 ; 01 = 0'3921. 

.-. ^ = 246 ; P=T l —T t = 147. 

Combining these equations T l is found to be equal to 246 lbs., = 
the maximum stress in the belt. Experiment shows that 70 lbs. 
per inch of width of a laced, single belt is a safe working stress. 
Therefore the width of the belt = 246 h- 70 = 3'5". The friction of 
the machine might have been estimated and added to the work to 
be done. 

84. Problem. — A sixty horse-power dynamo is to run 1500 revo- 
lutions per minute, and has a 15" pulley on its shaft. Power is 
supplied by a line shaft running 150 revolutions per minute. A 
suitable belt connection is to be designed. The ratio of angular 
velocities of dynamo shaft to line shaft is 10 to 1; hence the diame- 
ter of the pulley on the line shaft would have to be ten times as 
great as that of the one on the dynamo, == 12*5 feet, if the connec- 
tions were direct. This is inadmissible, and therefore the increase 
in speed must be obtained by means of an intermediate, or counter 
shaft. Suppose that the diameter of the largest pulley that can be 
used on the counter shaft = 48". Then the necessary speed of the 

15 
counter shaft = 1500 X ttt = 470 nearly. The ratio of diameters 

48 J 

of the required pulleys for connecting the line shaft and the counter 

470 
shaft = — ^ = 3" 13. Suppose that a 60" pulley can be used on the 

line shaft ; then the diameter of the required pulley for the counter, 

shaft will = 7— tt, = 19" nearly. Consider first the belt to connect 
3"13 J 



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86 MACHINE DESIGN. 

the dynamo to the counter shaft. The work = 60 X 33,000 = 
1,980,000 foot-pounds per minute ; the rim of the dynamo pulley 

7i1 5 

moves ~ X 1500 = 5890 feet per minute. Therefore T 1 —T 2 = 



1,980,000 



336 lbs. The axis of the counter shaft is 10 feet from 



5890 

VJU^ E — r 24 — 7*5 

the axis of the dynamo, and as before ^y* = — - — = — — — — = 

0-1378. Therefore = 8° nearly. 

a = 180° — 20 = 164° 
a in 7T measure = 164 X O0174 = 2*85. 

log - 1 = 0-4343 X 0-3 X 2'85 = 0*3713, 

* 2 

T 

| = 2-35. 

From these equations T x = 583 lbs., and the safe width of the 
single belting = 583 -s- 70 = 8'34"; say 8*5". The width of the belt 
to connect the line shaft to the counter shaft may be found by the 
same method. 

85. Table II is given to save the above calculations for each 
belt. The body of the table is made up of values of P, the driving 
force at the pulley surface. To use the table, suppose that the 
smaller angle of contact of the belt with the two pulleys considered 
is known, = a . Pis also known. Find a°, or the nearest smaller value, 
in the horizontal column at the head of the table. In the vertical 
column under this value of «, find P, or the next greater value. 
Horizontally opposite this in the first vertical column, is the safe 
width of a single belt. If a double belt is to be used the value found 
may be divided by 2. 

86. From equation (3), p. 82, it follows that the ratio of tensions, 

T 

~, when the belt slips at a certain allowable rate (i. e., when / is 

-* 2 

constant) depends only upon a. This ratio, therefore, is indepen- 



BELTS. 87 

dent of the initial tension, T. d ; hence "taking up" a belt does not 

T 

change -^. The difference of tension, T x — T 2 = P, is, however, de- 

pendent on T 3 . Because p, the normal pressure between belt and 
pulley, varies directly as T s . Then since dF = fpds — dT, it follows 
that dT varies with T,, and hence 



/ 



dT=T—T, = P 



varies with T. A . This is equivalent to saying that "taking up" a 
belt increases its driving capacity, and "letting it out" decreases 
its driving capacity. 

This result is modified because another variable enters the prob- 
lem. If jT 3 be changed, the amount of slipping changes, and the 
coefficient of friction varies directly with the amount of slipping. 
Therefore, an increase of T s would increase £) and decrease /in the 
expression fpds = dT, and the converse is also true. This is prob- 
ably of no practical importance. 

The value of P may also be increased by increasing either/, the 
coefficient of friction, or a, the arc of contact ; since increase of 

T 

either increases the ratio ~\ and therefore increases T x — T 2 = P. 

■L 2 

Increasing T A decreases the life of the belt. It also increases the 
pressure on the bearings in which the pulley shaft runs, and there- 
fore increases frictional resistance ; hence a greater amount of the 
energy supplied is converted into heat and lost to any useful pur- 
pose. But if T. A be kept constant and/ or « be increased, the driving 
power is increased without increase of pressure in the bearings, 
because this pressure = 2T 3 = constant. When possible, therefore, 
it is preferable to increase P by increase of/ or a, rather than by 
increase of T. 

Application of belt dressing may serve sometimes to increase/. 

87. If, as in Fig. 86, the arrangement is such that the upper 
side of the belt is the slack side, the "sag" of the belt tends to 

T 

increase the arc of contact, and therefore to increase — *. If the 



88 MACHINE DESIGN. 

lower side is the slack side, the belt sags away from the pulleys 

T 

and « and -^ are decreased. 

An idler pulley, C, may be used, as in Fig. 90. It is pressed against 

the belt by some means. Its purpose may be to increase P by 

T + T. 
increasing the tension, T s , = — — » — -. In this case friction in the 

bearings is increased. Or it may be used on a slack belt to increase 

T 

the angle of contact, «, the ratio -^, and therefore P, the driving 

T 4- T. 
force. In this case the value of T s , = - J -^ — 2 , may be made as 

small a value as is consistent with driving, and hence the journal 
friction may be small. 

Tighteners are sometimes used with slack belts for disengaging 
gear, the driving pulley being vertically below the follower. 

In the use of any device to increase/ and a, it should be remem- 
bered that T x is thereby increased, and may become greater than 
the value for which the belt was designed. This may result in 
injury to the belt. 

In Fig. 91, the smaller pulley, A, is above the larger one, B. A 
has a smaller arc of contact, and hence the belt would slip upon it 
sooner than on B. The weight of the belt, however, tends to increase 
the pressure between the belt and A, and to decrease the pressure 
between the belt and B. The driving capacity of A is thereby in- 
creased, while that of B is diminished ; or, in other words, the 
weight of the belt tends to equalize the inequality of driving power. 
If the larger pulley had been above, there would have been a ten- 
dency for the belt weight to increase the inequality of driving 
capacity of the pulleys. The conclusion from this, as to arrange- 
ment of pulleys, is obvious. 

88. A belt resists a force which tends to bend it. Work must be 
done, therefore, in bending a belt around a pulley. The more it is 
bent the more work is required. Suppose AB, Fig. 92, to represent 
a belt which moves from A toward B. If it runs upon C it must 



BELTS. 89 

be bent more than if it runs upon D. The work done in bending 
the belt is converted into useless heat by the friction between the 
belt fibres. It is desirable, therefore, to do as little bending as 
possible. This is one reason why large pulleys in general are more 
efficient than small ones. The resistance to bending increases with 
the thickness of the belt, and hence double belts should not be used 
on small pulleys if it can be avoided. 

89. Effect of Centrifugal Force of Belts. — In Fig. 93, as the belt 
reaches a, it has its direction of motion changed. The belt tends to 
move on in a straight line, and therefore resists the change of direc- 
tion. There results a force acting radially outward, which tends to 
cause the belt to leave the pulley. The measure of this force per 

w v 1 
linear inch of belt —c = — X — ; in which w = weight of belting per 

g r o ox- 

linear inch, v = belt velocity in feet per second, g = 32*2 feet per 
second acceleration, r = pulley radius. As the velocity of the belt 
is increased, w and r remaining constant, c will increase, and will 
eventually equal the radial pressure at a per linear inch of the belt 
= p. With further increase of v the belt would leave the pulley at 
a. This would result in a decrease in the arc of contact, and hence 
a decrease in the driving capacity of the belt. This centrifugal 
force is the same for every linear inch of the belt which is in con- 
tact with the pulley, i. e., the radial force acting outward is constant 
from a around to b. The radial pressure p, acting inward, however, 
increases from a around to b ; hence the tendency to leave the pulley 
is greatest at a. Experience shows that this may occur in practice, 
as shown in Fig. 94, the angle of contact being reduced from a to ft. 
The value of the radial pressure, = p, due to belt tension, at a, the 
middle of the first inch of contact, may be found. The value is less 
than for any other inch of contact, because it increases from a 
around to b. This value compared with the centrifugal force found 
as above shows the tendency for the belt to leave the pulley. 

To find p. — In Fig. 95 the first inch of belt in contact with the 
pulley at a, Fig. 93, is represented. This element of the belt sub- 
tends an angle 0, whose value depends on the radius of the pulley. 



90 MACHINE DESIGN. 

The element of the belt is in equilibrium under the influence of 
three forces, T 2 -f AT 2 , T, 2 -f- 4F, and the reaction of p; i. e., the pres- 
sure of the pulley against the element of the belt. This reaction — 
p, and its line of action (being radial) makes equal angles with the 
lines of action of T^ + ^T^and T 2 -\- dF. The angle between the 
lines of action of these equal forces (T 3 -\- dT a and T. z + AF) = 0. 
The force triangle is therefore the isosceles triangle shown. In 
this triangle 

p sinO 180° — , iJP T x frT, 

in which s = number of inches of contact of belt with pulley. 
Therefore 

T — T 

T,-\ — l - l sin 



V 



. 180° 
sin 



2 



T 

Apply this to the problem on page 84 : T x = 583 ; T 2 = — ^ = 247. 

Zoo 

^ — ^ = 336. 

s = aXr = 164° X 0-0174 X 7*5 = 21*4". S 

= angle subtended by 1" on 7*5" radius = 360° X - . = 7*64° 

Ji X {'ok 



7° 38'. Sin = 0-1328. 

180° — 172° 22' 



2 2 



= 86° 11': 



sin 1^^=0-9977; 

' p _( 247 + S)°- 1328 _, 62 . 7x 0.0 1 3 2 8_ 34 . 7 
P Mm - Ai 7 x "5^977 -d * 7. 

To find c. — A cubic inch of belting weighs about 0'04 lb. Single 
belting is about 0'25" thick, and in this case the width is 8-5". The 
weight of belt per linear inch is therefore W= 0*25 X 8*5 X 0*04= 
0-085 lbs. 



V = 



BELTS. 91 

1t) fift = 98*2 feet per second. 
1 'A X ou 

wv l 0-085 X 98-2 a 



'-" ' 82-2x1! 



40-8 lbs. 



The centrifugal force is in excess at (a) the middle of the first inch 
of contact, therefore, by an amount equal to 40'8 lbs. — 84' 7 lbs.= 
6*1 lbs. There would be a tendency for the belt to lift. This is 
opposed, however, by the weight of the belt. If the slack side of 
the belt be supposed to be straight and horizontal, one-half ^\^f"^—tf/ 
weight will be supported at a.^ The distance between centres of 
shafts = 10 ft. = 120". The weight W of the slack side = cubic con- 
tents X weight per cubic unit, = 0*25 X 8'5 X 120 X 0"04 = 10*2 lbs. 
= W. Half of this aids p in its opposition to c at a. Hence the out- 
ward radial force at a in this case= 408 — (34*7 -f 5*1) = 1 lb. If 
the direction of rotation were reversed the slack side would be below 
and the outward radial force at b would equal 40*8 — (34*7 — 5'1) 
= 11"2 lbs. If the line joining the centres is inclined, as in Fig. 
97, only the component of W at right angles to the belt, = P, is 
effective to produce inward radial pressure at a. If the slack side 
of the belt becomes vertical P becomes = 0, and hence the weight 
has no effect. 

To find the velocity of belt at which, under given conditions, 
the belt just tends to leave the pulley. — Let to = radial force at a 
due to belt weight. The belt just tends to leave the pulley when 
the sum of the inward radial forces = the sum of the outward radial 

Wv* 

forces ; or when c =p ±w: substituting value of c = , in which 

9 r 
W= weight of one lineal inch of the belt, and solving for v — 

j(p zt w) QT 

X — w ^ a ^ e ^ ^ enc ^ s t° leave the pulley, running under 

given conditions, it would seem that increasing the radius of the 
pulley upon which the slack side runs would reduce the centrifugal 

force, since c cc - ; but in order to keep the shaft running at the same 



92 MACHINE DESIGN. 

rate as before, the belt must run faster and c oc v' 2 . Also, since the 
moment to produce rotation is constant, the force (with the increase 
of lever arm due to increased size of pulley) is less, and hence 
(unless a wider belt than is necessary is used) the width of the belt 
and hence its w r eight per linear inch must be reduced, and c oc w. 

In the design of belting care should be taken not to make the 
distance betw r een the shafts carrying the pulleys too small, especially 
if there is the possibility of sudden changes of load. Belts have 
some elasticity, and the total yielding under any given stress is 
proportional to the length, the area of cross-section being the same. 
Therefore a long belt becomes a yielding part, or spring, and its 
yielding may reduce the stress due to a suddenly applied load to a 
safe value ; whereas in the case of a short belt, with other conditions 
exactly the same, the stress due to much less yielding might be 
sufficient to rupture or weaken the joint. 



CHAPTER VIII. 

DESIGN OF FLY-WHEELS. 

90. Often in machines there is capacity for uniform effort, but 
the resistance fluctuates. In other cases a fluctuating effort is 
applied to overcome a uniform resistance, and yet in both cases a 
more or less uniform rate of motion must be maintained. When 
this occurs, as has been explained, a moving body of considerable 
weight is interposed between effort and resistance, which, because of 
its weight, absorbs and stores up energy with increase of velocity 
when the effort is in excess, and gives it out with decrease of 
velocity when the resistance is in excess. This moving body is 
usually a rotating body, called a fly-wheel. 

To fulfill its office a fly-wheel must have a variation of velocity ; 
because it is by reason of this variation that it is able to store and 
give out energy. The kinetic energy, E, of a body whose weight is 
W, moving with a velocity v, is expressed by the equation 

£= V 

To change E, with W constant, v must vary. The allowable varia- 
tion of velocity depends upon the work to be accomplished. Thus, 
the variation in an engine running electric lights, or spinning 
machinery, should be very small ; probably not greater than a half 
of one per cent. While a pump or a punching machine may have 
a much greater variation without interfering with the desired re- 
sult. If the maximum velocity, v„ of the fly-wheel rim, and the 
allowable variation are known, the minimum velocity, v„ becomes 
known ; and the energy that can be stored and given out with the 



94 MACHINE DESIGN. 

allowable change of velocity is equal to the difference of kinetic 
energy at the two velocities. 

„ Wv* Wv* W, 2 , 

The general method for fly-wheel design is as follows : Find the 
maximum energy due to excess or deficiency of effort during a cycle 
of action, = E. Assume a convenient mean diameter of fly-wheel 
rim. From this and the given maximum rotative speed of the fly- 
wheel shaft, find i\. From v x and the given allowable variation of 
velocity, find v. r Solve the above equation for TT, thus : 

v* — v.* 

Substitute the values of E,v v v 2 , and g -- 32 - 2 ft. per second. Whence 
W becomes known, = weight of fly-wheel rim. The weight of rim 
only will be considered ; the other parts of the wheel being nearer 
the axis have less velocity, and less capacity per pound for storing 
energy. Their effect is to reduce slightly the allowable variation 
of velocity. 

91. Problem. — In a punching machine the belt is capable of 
applying a uniform torsional effort to the shaft ; but most of the 
time it is only required to drive the moving parts of the machine 
against frictional resistance. At intervals, however, the punch 
must be forced through metal which offers shearing resistance to its 
action. Either the belt or fly-wheel, or the two combined, must be 
capable of overcoming this resistance. A punch makes 30 strokes 
per minute, and enters the die \" . It is required to punch f" holes 
in steel plates \" thick. The shearing strength of the steel is about 
50,000 pounds per square inch. When the punch just touches the 
plate the surface which offers shearing resistance to its action 
equals the surface of the hole which results from the punching, == 
ndt, in which d = diameter of hole or punch, t = thickness of plate. 
The maximum shearing resistance, therefore, equals tt| X -J X 50000 
= 58800 lbs. As the punch advances through the plate the resist- 
ance decreases, because the surface in shear decreases, and when the 
punch just passes through the resistance becomes zero. If the 



DESIGN OF FLY-WHEELS. 95 

change of resistance be assumed uniform (which would probably be 

approximately true) the mean resistance to punching would equal 

., . ' . . . . „ 58800 + 
the maximum resistance -J- minimum resistance, -"- z, = 

= 29400. The radius of the crank which actuates the punch = 2 ". 
In Fig. 98 the circle represents the path of the crank-pin centre. 
Its vertical diameter then represents the travel of the punch. If 
the actuating mechanism be a slotted cross-head, as is usual, it is 
a case of harmonic motion, and it may be assumed that while the 
punch travels vertically from A to B, the crank-pin centre travels 
in the semicircle ACB. Let BD and DE each = -J inch. Then 
when the punch reaches E it just touches the plate to be punched, 
which is \" thick, and when it reaches D it has just passed through 
the plate. Draw the horizontal lines EF and DG and the radial 
lines OG and OF. Then, while the punch passes through the plate, 
the crank-pin centre moves from F to G, or through an angle (in 
this case) of 19°. Therefore the crank shaft A, Fig. 99, and attached 
gear rotate through 19° during the action of the punch. The ratio 
of angular velocity of the pinion and the gear = the inverse ratio of 

ah 
pitch diameters = — = 5. Hence the shaft B rotates through an 

angle = 19° X 5 = 95° during the action of the punch. If there 
were no fly-wheel the belt would need to be designed to overcome 
the maximum resistance ; i. e., the resistance at the instant when 
the punch is just beginning to act. This would give for this case a 
double belt about 20" wide. The need for a fly-wheel is therefore 
apparent. Assume that the fly-wheel may be conveniently 36" 
mean diameter, and that a single belt 5" wide is to be used. The allow- 
able maximum tension is then-=5 X allowable tension per inch of 
width of single belting = 5 X 70 = 350 lbs.= T v Then from the 

T T T S50 

equation =f = eS", if «° = 180°, ^ =2-56; hence T.=^- = ^— = 

136*5, and 7\— T. 2 = 213*5 lbs. = the driving force at the surface 
of the pulley. Assume that the frictional resistance of the machine 
is equivalent to 25 lbs. applied at the pulley rim. Then the belt 
can exert 213'5 lbs. — 25 = 188*5 lbs., = P, to accelerate the fly-wheel 



96 MACHINE DESIGN. 

or to do the work of punching. Assume variation of velocity--- 10 
per cent. The work of punching = the mean resistance offered to 
the punch multiplied by the space through which the punch acts, = 

^|52 x o-5 = 14700 inch-pounds = 1220 foot-pounds. The pulley 

shaft moves during the punching through 95°, and the driving ten- 
sion of the belt, = P= 188*5 lbs., does work = P X space moved 

95° 
through during the punching = 188*5 lbs. X *d ^7777 — 188*5 lbs. X * 

95 
X 2 ft. X — = 311 foot-pounds. The work left for the fly-wheel 

to give out with a reduction of velocity of 10 per cent. = 1220 — 311 
= 909 foot-pounds. Let i\ = maximum velocity of fly-wheel rim ; 
v. 2 = minimum velocity of fly-wheel rim ; W— weight of the fly- 
wheel rim. The energy it is capable of giving out, while its velocity 

W(v 2 — v. 2 ) 
is reduced from i\ to v 2J = — ^-^. — , and the value of Tl^must be 

such that this energy given out shall equal 909 foot-pounds. Hence 
the following equation may be written : 

Therefore r = 909. X 2 X <, 

I?! 2 V./ 

The punch shaft makes 30 revolutions per minute and the pulley 

shaft 30 X 5 = 150 = N revolutions per minute. Hence v x in feet 

Nd* 
per second = -^- ; d being fly-wheel diameter in feet, 

150 X 3* 
*i = — 60- =235. 

^ = 0-90^ = 21-1. 

v * = 552 ; v. 2 = 446 ; v 2 — v 2 = 106. 

rj w , 909 X 2 X 32*2 

Hence W= t— ^ = 551 lbs. 

lUb 



DESIGN OF FLY-WHEELS. 97 

To proportion the rim. — A cubic inch of cast iron weighs 0*26 lbs.; 

551 
hence there must be =— ^ = 2120 cu. in. The cubic contents of the 
(J '2b 

rim = mean diameter X * X its cross-sectional area. A, — 2120 cu. 

, A 2120 : 

m. ; hence A -— ^^ — lo 8 sq. in. 

36' X * H 

If the cross-section were made square its side would = -|X'18"8 — 
4-34". 

92. Pump Fly-Wheel. — The belt for the pump, p. 83, is designed 
for the average work. A fly-wheel is necessary to adapt the vary- 
ing resistance to the capacity of the belt. The rate of doing work 
on the return stroke (supposing no resistance due to suction) is 
only equal to the frictional resistance of the machine. During the 
working stroke the rate of doing work varies because the velocity of 
the plunger varies, although the pressure is constant. The rate of 
doing work is a maximum when the velocity of the plunger is 
greatest. In Fig. 100, A is the velocity diagram ; B is the force 
diagram ; C is the tangential diagram drawn as indicated on pages 
35-36. The belt, 3*5" wide, is capable of applying a tangential 
force of 147 lbs. to the 18" pulley rim. The velocity of the pulley 
rim = - 1*5 X 300 = 1410. The velocity of the crank-pin axis = 
- 0-833 X 50 = 130*8. Therefore the force of 147 lbs. at the pulley 

rim corresponds to a force = 147 X iPf . Q == 1585 lbs. applied tan- 

gentially at the crank-pin axis. This may be plotted as an 
ordinate upon the tangential diagram C, from the base line XX U 
using the same force scale. Through the upper extremity of this 
ordinate draw the horizontal line DE. The area between DE and 
XX l represents the work the belt is capable of doing during the 
working stroke. During the return stroke it is capable of doing the 
same amount of work. But this work must now be absorbed in 
accelerating the fly-wheel. Suppose the plunger to be moving in 
the direction shown by the arrow. From E to F the effort is in 
excess and the fly-wheel is storing energy. From F to G the resist- 



98 MACHINE DESIGN. 

ance is in excess and the fly-wheel is giving out energy. The work 
the fly-wheel must be capable of giving out with the allowable 
reduction of velocity is that represented by the area under the curve 
above the line FG. From G to D, and during the entire return 
stroke, the belt is doing work to accelerate the fly-wheel. This work 
becomes stored kinetic energy in the fly-wheel. Obviously the 
following equation of areas may be written : 

XJEF + XGD + XHKX, == GMF. 

The left hand member of this equation represents the work done by 
the belt in accelerating the fly-wheel ; the right hand member 
represents the work given out by the fly-wheel to help the belt. 

The work in foot-pounds represented by the area GMF may be 
equated with the difference of kinetic energy of the fly-wheel at 
maximum and minimum velocities. To find the value of this 
work : One inch of ordinate on the force diagram represents 4260 
lbs. ; one inch of abscissa represents 0*2245 ft. Therefore one 
square inch of area represents 4260 lbs. X 02245 = 956*37 foot- 
pounds. The area GMF=1'Q sq. in. Therefore the work = 
956-37 X 1*6 = 1530 foot-pounds = E. The difference of kinetic 

W 

energy = — (v x 2 — v./) = 1530 ; W equals the weight of the fly- 
wheel rim. Hence 

1530 X 32'2 



W = 



v./ 



Assume the mean fly-wheel diameter = 2*5 ft. It will be keyed to 
the pulley shaft, and will run 300 revolutions per minute, = 5 
revolutions per second. The maximum velocity of fly-wheel rim ■== 
2*5~ X 5 — 39*15 = v v Assume an allowable variation of velocity, 
= 5 per cent. Then v 2 = 39*15 X 0*95 = 37*19 ; v* = 1532*7 ; 
v* = 1383*1 ; v t * — v./ = 149*6. Hence 

w= 1530 X 32-2 =3291bg 
i4y*o 




1L 



m./oo . 



ff 



gear 




$ 



:„i05^_JC 



i^K 




■s-^ 



f/y ™hee /. 



c/r/v/'rw 



pu/ley. 



I /V4 . 0$ 



DESIGN OF FLY-WHEELS. 99 

There must be 829 ~- 026 cubic inches in the rim = 1262. The 
pitch circumference — 30 X ~ = 94*2". Hence the cross-sectional 
area of rim = 1262 -s- 94 - 2 = 13*4 -f- . The rim may be made 
3" X 4-5". 

The frictional resistance of the machine is neglected. It might 
have been estimated and introduced into the problem as a constant 
resistance. 

93. Steam Engine Fly-Wheel. — From given data draw the indi- 
cator card as modified by the acceleration of reciprocating parts. 
See page 35 and Fig. 30. From this, and the velocity diagram, 
construct the diagram of tangential driving force, Fig. 31. Measure 
the area of this diagram and draw the equivalent rectangle on the 
same base. This rectangle represents the energy of the uniform 
resistance during one stroke ; while the tangential diagram repre- 
sents the work done by the steam upon the crank-pin. The area 
of the tangential diagram which extends above the rectangle repre- 
sents the work to be absorbed by the fly-wheel with the allowable 
variation of velocity. Find the value of this in foot-pounds, and 
equate it to the expression for difference of kinetic energy at maxi- 
mum and minimum velocity. Solve for W 3 the weight of fly-wheel. 



CHAPTER IX. 

RIVETED JOINTS. 

94. A rivet is a fastening used to unite metal plates or rolled 
structural forms, as in boilers, tanks, built-up machine frames, etc. 
It consists of a head, A, Fig. 101, and a straight shank, B. It is in- 
serted, usually red-hot, into holes, either drilled or punched in the 
parts to be connected, and the projecting end of the shank is then 
formed into a head (see dotted lines) either by hand or machine 
riveting. A rivet is a permanent fastening and can only be removed 
by cutting off the head. A row of rivets joining two members is 
called a riveted joint or seam, of rivets. In hand riveting the project- 
ing end of the shank is struck a quick succession of blows with 
hand hammers and formed into a head by the workman. A helper 
holds a sledge or "dolly bar" against the head of the rivet. In 
"button set" or "snap" riveting, the rivet is struck a few heavy 
blows with a sledge to "upset" it. Then a die or "button set," 
Fig. 102, is held with the spherical depression, B, upon the rivet ; 
the head A is struck with the sledge, and the rivet head is thus 
formed. In machine riveting a die similar to B is held firmly in 
the machine and a similar die opposite to it is attached to the pis- 
ton of a steam, hydraulic, or pneumatic cylinder. A rivet, properly 
placed in holes in the members to be connected, is put between the 
dies and pressure is applied to the piston. The movable die is 
forced forward and a head formed on the rivet. 

The relative merits of machine and hand riveting have been 
much discussed. Either method carefully carried out will produce 
a good serviceable joint. If in hand riveting the first few blows be 
light the rivet will not be properly upset, the shank will be loose in 



RIVETED JOINTS. 101 

the hole, and a leaky rivet results. If in machine riveting the axis 
of the rivet does not coincide with the axis of the dies, an off-set 
head results. See Fig. 103. In large shops where work must be 
turned out economically in large quantities, machines must be used. 
But there are always places inaccessible to machines, where the 
rivets must be driven by hand. Holes for the reception of rivets 
are usually punched, although for thick plates and very careful 
work they may be sometimes drilled. If a row of holes be punched 
in a plate, and a similar row as to size and spacing be drilled in the 
same plate, testing to rupture will show that the punched plate is 
weaker than the drilled one. If the punched plate had been 
annealed it would have been nearly restored to the strength of the 
drilled one. If the holes had been punched i" small in diameter 
and reamed to size, the plate would have been as strong as the 
drilled one. These facts, which have been experimentally deter- 
mined, point to the following conclusions : First, punching injures 
the material and produces weakness. Second, the injury is due to 
stresses caused by the severe action of the punch, since annealing, 
which furnishes opportunity for equalization of stress, restores the 
strength. Third, the injury is only in the immediate vicinity of 
the punched hole, since reaming out V on a side removes all the 
injured material. In ordinary boiler work the plates are simply 
punched and riveted. If better work is required the plates must be 
drilled, or punched small and reamed, or punched and annealed. 
Drilling is slow and therefore expensive : annealing is apt to change 
the plates and requires large expensive furnaces. Punching small 
and reaming, is probably the best method. 

95. Riveted Joints are of two general kinds : First, Lap Joints, 
in which the sheets to be joined are lapped upon each other and 
joined by a seam of rivets, as in Fig. 104 a. Second, Butt Joints, 
in which the edges of the sheets abut against each other, and a 
strip called a "cover plate" or "butt strap" is riveted to both 
edges of the sheet, as in c. 

There are two kinds of riveting : Single, in which there is but 
one row of rivets, as in a ; and double, where there are two rows. 



102 MACHINE DESIGN. 

Double riveting is subdivided into "chain riveting," b, and u zig- 
zag" or "staggered" riveting, d. 

Lap joints may be single, double chain, or double staggered 
riveted. 

Butt joints may have a single strap, as in c, or double strap ; 
i. e., an exactly similar one is placed on the other side of the joint. 
Butt joints with either single or double strap may be single, double 
chain, or double staggered riveted. 

To sum up, there are : 

fSingle Riveted 

Lap Joints^ Double Chain u 

I " Staggered 

f ( Single Riveted 

j Single Straps Double Chain 

^ " Staggered " 
Butt Joints<( 

fSingle Riveted 

I Double " << Double Chain 
L I " Staggered 

A riveted joint may yield in any one of four ways : First, by 
the rivet shearing. Second, by the plate yielding to tension on the 
line AB, Fig. 105 a. Third, by the rivet tearing out through the 
margin, as in c. Fourth, the rivet and sheet bear upon each other 
at D and E in d, and are both in compression. If the unit stress 
upon these surfaces becomes too great, the rivet is weakened to 
resist shearing, or the plate to resist tension, and failure may occur. 
This pressure of the rivet on the sheet is called " bearing pressure." 

96. As a preliminary to the designing of joints it is necessary to 
know the strength of the rivets to resist shear ; of the plate to resist 
tension ; and of the rivets and plate to resist bearing pressure. 
These values must not be taken from tables of the strength of 
the materials of which the plate and rivets are made, but must be 
derived from experiments upon actual riveted joints tested to rup- 
ture. The reason for this is that the conditions of stress are modi- 




u 



r/#./03. 




F/Q./02 . 





uM^&A 





rttfpQ^ ezzzz^fgg^ 23233 



RIVETED JOINTS. 



108 



fied somewhat in the joint. For instance, in single strap butt 
joints, and in lap joints, the line of stress being the centre line of 
plates, and the plates joined being offset, flexure results and the 
plate is weaker to resist tension ; if the joint yield to this stress in 
the slightest degree the " bearing pressure" is localized, and becomes 
more destructive. Extensive and accurate experiments have been 
made upon actual joints and the results are available in Stoney's 
" Strength and Proportions of Riveted Joints." The constants 
given are taken from this book. 



Steel. 




Ultimate shearing strength of rivets, single shear 

double " 

Ultimate tensile strength of plate between rivet holes, 

single shear 

Ultimate bearing pressure per sq. inch of diametral 

plane of rivet, single shear 

Ultimate bearing pressure per sq. inch of diametral 

plane of rivet, double shear 



97. The theoretical diameter of rivet for a given thickness of 
plate may now be determined. Let d — diameter of rivet hole ; t = 
thickness of plate; p = pitch of rivets; T '= ultimate tensile strength 
of plate between rivet holes ; £ = ultimate shearing strength of 
rivets ; C = ultimate bearing pressure. 

The strength of the rivet to resist shearing at AB, Fig. 106, 
should be equal to its strength to resist bearing pressure at A and 0, 
and hence the expressions for those strengths may be equated, thus: 



Solving, 



Ctd 

Ct 



4 



67000 



'8 X 0-7854 40000 X 0'7854 



t = 2«. 



Hence, for equal strength to resist bearing pressure and shear, the 
diameter of the rivet should equal twice the thickness of the plate. 



104 



MACHINE DESIGN. 



Let the results thus derived be compared with the values that are 
used in actual practice. See table. 

Comparative Values in Inches of Rivet Diameter for Different Values 
of Thickness of Plate. 



t 


2/ 


l.2 X /t 


d 


3 x6 


K 





% 


H . 


K 


9 i6 


% 


5 i6 


5 A 


"X6 


% 


% 


X 


% 


%-K 


% 


1 


7 A 


%-% 


% 


1M 


I5 !6 


%-i 


% 


IK 


l X i6 


i-iM 


% 


m 


w 


i-ijtf 


1 


2 


l 3 x6 


l-i % 


m 


2M 





1X-W 



The first column gives the thickness of the plate ; the second the 
diameter of the rivet = 2t; the third gives the rivet diameter calcu- 
lated from the formula of Professor Unwin, d = 1*2 \/t\ the fourth 
column gives rivet diameters as found in practice, taken from 
Stoney's book, page 12. d — 2t agrees with practice up to ■§ " plates, 
but for thicker plates it gives values that are too large. The reason 
for this is that the difficulty in driving rivets increases very rapidly 
with their size ; 1^" or If" being the largest rivet that can be driven 
conveniently. The equality of strength to resist bearing pressure 
and shear is therefore sacrificed to convenience in manipulation. 
As the diameter of the rivet is increased the area to resist bearing 
pressure increases less rapidly than the area to resist shear (the 
thickness of the plate remaining the same), the former varying as 



RIVETED JOINTS. 105 

d, and the latter as d'\ therefore if d is not increased as much as is 
necessaiy for equality of strength, the excess of strength will be 
to resist bearing pressure. If the other parts of the joint are made 
as strong as the rivet in shear, and this strength is calculated from 
the stress to be resisted, the joint will evidently be correctly pro- 
portioned. : 

To calculate the diameter of rivet for a butt joint with double 
cover plates. — The rivet is in double shear, and therefore ultimate 
bearing pressure = 89000 lbs. per square inch = C. And also ulti- 
mate shear pressure = 35000 lbs. per square inch = S'. 

Equating as before Cat = — j — — . 

_, ,,, , 2Ct 2 X 89000 X t 1 . ft# . 

From which a = ^- = . nm — = 16 t nearly. 

Comparison of results of this formula with tables of dimensions of 
practice, shows them to be too large. The following empirical for- 
mulas may be trusted : 

For thin plates — for iron d = l'St; for steel, d —- 1*25 t. 
" thick " " d = l'lt; " d = 1-125*. 

98. The next value to be determined is the pitch of the rivets, 
i. e., the distance from the centre of one rivet to the centre of the 
next one. See Fig. 107. It is required to make the pitch of such a 
value that the strength of the plate between rivet holes to resist 
tension shall equal the strength of the rivet to resist shear. It has 
already been shown that the strength to resist bearing pressure is 
equal to, or greater than, the strength to resist shear. Equate ex- 
pressions for shearing strength of the rivet, and tensile strength of 
the plate on a section through the rivet hoies, and solve for p = 
pitch. For a single riveted lap joint, 

7 ^S=Tt(p-d). 

„ , . , 0'7854r^S Y + Ttd 

t rom which p = — . 

14 it 



106 MACHINE DESIGN. 

Let S = 40000 and T= 40000. 

Then if t = i", d = |"; p = 1.28". 

t = r, d= f ; p = 1.79". 

t = r, d= F? p = 2M". 

t = l", d = H"\ p = 2.l2". 

All of these agree with Stoney's "Table of Boilermaker's Propor- 
tions," lap joints, iron plates, and rivets, except for t = i". This 
formula may, therefore, be trusted except for very thin plates. 

To find p for butt joints with double straps, single riveted. — 
Since the rivet is in double shear, 

2 X 0'7854#flr + Ttd 
*'= Tt ; 

S' = 35000 lbs. per square inch, the value for double shear. 

For steel plates and steel rivets, the values of the constants, T 
and S, need to be changed in above formulas. See values given. 

To rind the pitch of double riveted joints the method is the same. 
There are, however, two rivets now to support the strip of plate 
between holes, instead of one, as in the single joint. See Fig. 107. 
Therefore the first formula for p, multiplying the shearing strength 
by 2, becomes 

l'57d 2 S + Ttd 
* = Tt • 

„ , , , , 3'Ud*S' + Ttd 
r or double shear p = ■= — - — — ; 

S' being value for double shear. 

99. The margin in a riveted joint is the distance from the edge 
of the sheet to the rivet hole. This must be made of such value 
that there shall be safety against failure by the rivet tearing out. 
There can be no satisfactory theoretical determination of this value ; 
but practice and experiments with actual joints show that a joint 
will not yield in this way if the margin be made = d = diameter of 
the rivet. The distance between the centre lines of the rows in 
double chain riveting may be taken = 2'5 d ; and in double stag- 



RIVETED JOINTS. 



107 



gered riveting may be taken = l'88d. Thus the total width of lap 
for single riveting equals 3d; in double chain riveting = 5'5d; and in 
double staggered riveting 4*88d. The riveted joints considered can- 
not be as strong as the unperforated plates. The ratio of strength 
of joint to strength of plate is called joint efficiency. If the joint 
were equally strong to resist rupture in all possible ways, the joint 
efficiency would equal the ratio of area of plate through rivet sec- 
tion, to the area of unperforated section. Results obtained in this 
way differ somewhat from the results of actual tests, and the latter 
values should be used. See following tables. 

Relative Efficiency of Iron Joints. 



Efficiency 
Per Cent. 



Original solid plate 

Lap Joint, single riveted, punched. . . . 

" " " drilled 

double " 

Butt Joint, single cover, single riveted 
" double " 
double " single " 
" double " 



100 
45 
50 
60 
45-50 
60 
55 
66 



Relative Efficiency of Steel Joints. 



Original solid plate 

Lap Joint, single riveted, punched 

drilled 

double " punched 

drilled 

Butt Joint, double cover, double riveted, punched. 

" " drilled... 



Efficiency Per Cent. 



^Thickness of Plates.—^ 



100 
50 
55 
75 
80 
75 
80 



1^-5/ 



100 
45 
50 
70 

75 
70 

75 



100 
40 
45 
65 
70 
65 
70 



These tables are from Stoney's " Strength and Proportions of 
Riveted Joints." 



108 MACHINE DESIGN. 

100. The following problem will serve to illustrate the design of 
riveted joints for boilers. It is required to design a horizontal 
tubular boiler 48" diameter to carry a working pressure of 100 lbs. 
per square inch. A boiler of this type consists of a cylindrical 
shell of wrought iron or steel plates made up in length of two or 
more courses or sections. Each course is made by rolling a flat 
sheet into a hollow cylinder and joining its edges by means of a 
riveted joint, called the longitudinal joint or seam. The courses are 
joined to each other also by riveted joints, called circular joints or 
seams. Circular heads of the same material have a flange turned 
all around their circumference, by means of which they are riveted 
to the shell. The proper thickness of plate may be determined 
from : I. The diameter of shell = 48". II. The working steam 
pressure per square inch = 100 lbs. III. The tensile strength of 
the material used ; let steel plates be used of 60000 lbs. specified 
tensile strength. Preliminary investigation of the conditions of 
stress in the cross-section of material cut by a plane. — I. Through 
the axis ; II. At right angles to the axis, of a thin hollow cylinder ; 
the stress being due to the excess of internal pressure per square 
inch over the external pressure per square inch. Let l = ihe length 
of the cylindrical shell in inches ; d = the diameter of the cylin- 
drical shell in inches ; p = the excess of internal over external 
pressure per square inch ; i^ — unit stress in a longitudinal section 
of material of the shell due to p; p 2 = unit stress in a circular sec- 
tion of material of the shell due to p; t — thickness of plate; T= 
ultimate tensile strength of plate. 

In a longitudinal section the stress = Idp, and the area of metal 

d lo 

sustaining it = 2lt. Then p l — ~. 

At 

Tcd^T) 

In a circular section the stress = — 7^-, and the area = ~dt nearly. 

rn, ~d 2 p w 1 dp 

Therefore the stress in the first case is twice as great as in the 
second ; and a thin hollow cylinder is twice as strong to resist rup- 



RIVETED JOINTS. 109 

ture on a circular section as on a longitudinal one. The latter 
only, therefore, need be considered in determining the thickness of 
plate. Equating the stress due to p in a longitudinal section and 
the strength of the cross-section of plate that sustains it, we have 

ldp = 2UT. Therefore t = -^ = the thickness of plate that would 

just yield to the unit pressure p. To get safe thickness, a factor of 
safety must be used. It is usually equal in boiler shells to 4, 5, or 
6. Its value is small because the material is highly resilient and 
the changes of pressure are gradual, i. e., there are no shocks. This 
takes no account of the riveted joint, which is the weakest longi- 
tudinal section, e times as strong as the solid plate ; e being the 
joint efficiency, = 0'7o if the joint be double riveted. The formula 

then becomes t = ^ . Substituting values 

6X48X100 nQ0 „ 

'= 2 X 60000 X 0-75 = °' 32 ' Say * l6 ' 

The circular joints will be single riveted and joint efficiency will — 
0'50. But the stress is only one-half as great as in the longitudinal 
joint, and therefore it is stronger in the proportion 0*50 X 2 to 
0*75 = 1 to 0'75. From this it is seen that a circular joint whose 
efficiency is O'oO is as strong as the solid plate in a longitudinal 
section. From the value of t the joints may now be designed. 
Diameter of rivet = d = l^j/t =. l'2-|/0-3125 = 0'672", say 0'687" = 
ll 1 e. The pitch for a single riveted joint = 

0'7854d 2 £ + Ttd 
?= Tt ; 

But ^ = ^6 = 687"; £ = 50000 for steel; T= 60000 for steel; * = 
s l6 — 0"3125. Substituting these values _p = l'42". For double 
riveted joint 

p = ^ = (substituting as above) 2*66 . 



110 MACHINE DESIGN. 

The margin ==d = 0*687" = xl l6 ". The longitudinal joint will be 
staggered riveted and the distance between rows = l'SSd = 1*29" = 
say 1 5 !6". The total lap in the longitudinal joint — 4 - 88d = 3*35". 
The total lap in the circular joint — 3d = 2 l I e". The joints are 
therefore completely determined, and a detail of each, giving dimen- 
sions, may be drawn for the use of the workmen who make the tem- 
plets and lay out the sheets. 



CHAPTER X. 



DESIGN OF JOURNALS. 



101. Journals and the bearings or boxes with which they engage 
are the elements used to constrain motion of rotation or vibration 
about axes in machines. Journals are usually cylindrical, but may 
be conical, or, in rare cases, spherical. The design of journals, as 
far as size is concerned, is dictated by one or all of the three follow- 
ing considerations : I. To provide for safety against rupture or 
excessive yielding under the applied forces. II. To provide for 
maintenance of form. III. To provide against the squeezing out 
of the lubricant. To illustrate I. — Let Fig. 108 represent a pulley 
on the end of an overhanging shaft, driven by a belt, ABC. Rota- 
tion is as indicated by the arrow, and the belt tensions are T l and 
T 2 . The journal, J, engages with a box or bearing, D. The follow- 
ing stresses are induced in the journal : Torsion, measured by the 
torsional moment ( ^ — T t )r. Flexure, measured by the bending 
moment (T x -\- T 2 )a. This assumes a rigid shaft or a self-adjusting 
box. Shear, resulting from the force T x -f T. z . This journal must 
therefore be so designed that rupture or undue yielding shall not 
result from any one of these stresses. To illustrate II. — Consider 
the spindle journals of a grinding lathe. The forces applied are 
very small ; but the form of the journals must be maintained to 
insure accuracy in the product of the machine. A relatively large 
wearing surface is therefore necessary. To illustrate III. — The pres- 
sure upon a journal resulting from the applied forces may be 
sufficiently great to squeeze out the lubricant. Metallic contact, 
heating, and abrasion of the surfaces would result. In what fol- 



112 MACHINE DESIGN. 

lows, the area of a journal means its projected area ; i. e., its length 
X its diameter. 

The allowable pressure per square inch of area of a journal 
varies with several conditions. To make this clear, suppose a drop 
of oil to be put in the middle of an accurately finished surface 
plate ; suppose another exactly similar plate to be placed upon it 
for an instant ; the oil drop will be spread out because of the force 
due to the weight of the upper plate. If the plate were allowed to 
remain a longer time, the oil would be still further spread out, and 
if its weight and the time were sufficient, the oil would finally be 
entirely squeezed out from between the plates, and the metal sur- 
faces would come in contact. The squeezing out of the oil from 
between the rubbing surfaces of a journal and its box is, therefore, 
a function of the time as well as of pressure. If the surfaces under 
pressure move over each other, the removal of the oil is facilitated. 
The greater the velocity of movement, the more rapidly will the oil 
be removed, and therefore the squeezing out of the oil is also a 
function of the velocity of rubbing surfaces. 

When a journal is subjected to continuous pressure in one direc- 
tion, as for instance in a shaft with a constant belt pull, or with a 
heavy fly-wheel upon it, this pressure has sufficient time to act, and 
is therefore effective for the removal of the oil. But if the direction 
of the pressure is periodically reversed, as in the crank-pin of a 
steam engine, the time of action is less, the tendency to remove the 
oil is reduced, and the oil has opportunity to return between the 
surfaces. Hence, a higher pressure per square inch of journal 
would be allowable in the second case than in the first. 

If the direction of motion is also reversed, as in the cross head 
pin of a steam engine, the oil not only has an opportunity to return 
between the surfaces, but is assisted in doing so by the reversed 
motion. Therefore, a still higher pressure per square inch of jour- 
nal is allowable. Practical experience bears out these conclusions. 
Thus in journals with the direction of pressure constant, it is found 
that with ordinary conditions of lubrication the heating and "seiz- 
ing " or " cutting " occur quickly if the pressure per square inch of 



DESIGN OF JOURNALS. 113 

journal exceed about 380 lbs.* But in the crank-pins of punching 
machines, where the pressure acts for an instant, with quite an 
interval of rest, and where the velocity of rubbing surface is very 
low indeed, the pressure is often as high as from 2000 to 3000 lbs., 
per square inch, and there is no tendency to heating or abrasion. 
In engine crank-pins the pressure may be from 400 to 800 lbs. 
depending on the velocity of rubbing surface, and in cross-head 
pins where the velocity is always low it may be from 600 to 1000 
lbs. The value to be used in each particular case must be decided 
by the judgment of the designer. 

But even if the conditions are such that the lubricant is retained 
between the rubbing surfaces, heating may occur. There is always 
a frictional resistance at the surface of the journal ; this resistance 
may be reduced : a, by insuring accuracy of form and perfection of 
surface in the journal and its bearings ; 6, by insuring that the 
journal and its bearing are in contact, except for the film of oil, 
throughout their entire surface, by means of rigidity of framing or 
self-adjusting boxes, as the case may demand ; c, by selecting a suit- 
able lubricant to meet the conditions, and maintaining the supply to 
the bearing surfaces. By these means the friction may be reduced 
to a very low value, but it cannot be reduced to zero. 

There must be some frictional resistance,.and it is always con- 
verting mechanical energy into heat. This heat raises the tempera- 
ture of the journal and its bearing. If the heat thus generated is 
conducted and radiated away as fast as it is generated, the box 
remains at a constant low temperature. If, however, the heat is 
generated faster than it can be disposed of, the temperature of the 
box rises till its capacity to radiate heat is increased by the in- 
creased difference of temperature of the box and the surrounding 
air, so that it is able to dispose of the heat as fast as it is generated. 
This temperature, necessary to establish the equilibrium of heat 
generation and disposal, may under certain conditions be high 
enough to destroy the lubricant, or even to melt out a babbitt metal 

*See Mr. Tower's experiments in the "Minutes of the Institution of 
Mechanical Engineers." 



114 MACHINE DESIGN. 

box lining. Suppose now that a journal is running under certain 
conditions of pressure and surface velocity, and that it remains 
entirely cool. Suppose next that while all other conditions are 
kept exactly the same, the velocity is increased. All modern 
experiments on the friction in journals show that the friction in- 
creases with the increase of the velocity of rubbing surface. Therefore 
the increase in velocity would increase the frictional resistance at 
the surface of the journal, and the space through which this resist- 
ance acts would be greater in proportion to the increase in velocity. 
The work of the friction at the surface of the journal is therefore 
increased, because both the force and the space factors are increased. 
It is this work of friction which has been so increased, that produces 
the heat that tends to raise the temperature of the journal and its 
box. The rate of generation of heat has therefore been increased 
by the increase in velocity, but the box has not been changed in 
any way and therefore its capacity for disposing of heat is the same 
as it was before, and hence the tendency of the journal and its 
bearing to heat is greater than it was before the increase in velocity. 
Some change in the proportions of the journal must be made in 
order to keep the tendency to heat the same as it was before the 
increase in velocity. If the diameter of the journal be increased, 
the radiating surface of the box will be proportionately increased. 
But the space factor of the friction will be increased in the same 
proportion, and therefore it will be apparent that this change has 
not affected the relation of the rate of generation of heat to the dis- 
posal of it. But if the length of the journal be increased, the work 
of friction is the same as before and the radiating surface of the 
box is increased and the tendency of the box to heat is reduced. If, 
therefore, the conditions are such that the tendency to heat in a 
journal, because of the work of the friction at its surface, is the 
vital point in design, it will be clear that the length of the journal 
is dictated by it, but not the diameter. The reason why high speed 
journals have greater length in proportion to their diameter than 
low speed journals will now be apparent. 

102. Problem. — To design the main journal of a side crank 



DESIGN OF JOURNALS. 115 

engine. — The data are as follows: Diameter of steam cylinder = 
16" ; boiler pressure = 100 lbs. per sq. in. by gauge. Then the 
maximum force upon the steam piston due to steam pressure = 
100 X 8' 2 ~ — 20106 lbs. Suppose that the least expensive stress 
member is a "breaking piece," i. e., it will yield and relieve the 
stress in the other stress members when the total applied force = 
80000 lbs., about four times the maximum working force. In 
Fig. 109, DE is the centre line of the engine ; C is the crank- 
pin ; A is the crank disc, and B is the journal to be designed. 
The force P, = 80000 lbs., is the greatest force that can act in 
the line DE. The journal is supported up to the line FG. . 
In the section HK there is flexure stress measured by the flex- 
ure moment PI. I in this case == 6". The breaking piece only 
yields when the crank is at or near its centre ; hence, the 
torsional stress may be neglected. The radius of the shaft for 
safety against the moment PI, may be found from the formula 

SI Pic -r* 

PI = — ; from which /.= -—-. But / for circular section = — - ; 
c S 4 ? 

4PI 
and c = r. Hence 'r 3 = r— — . Let S for machinery steel = 12000. 

This gives a factor of safety = t^t™ = 5. Substituting values, 

P= 80000, I — 6", and S = 12000, in the above equation, gives r = 
8*7 1", say 3f". Hence, the shaft diameter = 7-J". This value de- 
pends upon the assumptions made for P, the strength of the break- 
ing piece, and for S, the safe stress for the material used. Different 
values might have been assumed, and would, of course, have given 
different results. The length of such a journal is determined by 
practical considerations. In this case the length should be about 
twice the diameter = 15", in order that convenient means may be 
supplied for taking up wear. The projected area of journal = 7*5* 
X 15 = 112*5 square inches. Assume 350 lbs. safe pressure per 
square inch of journal. This would admit of a working pressure 
of 850 X 112*5 = 39375. It is evident without investigation that 
this is greater than any working load for this journal. 



116 MACHINE DESIGN. 

103. Problem. — To design the crank-pin for the same engine. 

— The bending moment now equals Pl Y . Assume l x = 3". Then 

. 4 X 80000 X3, , . , QA „ _, , , 

r 6 = . ^^ — , from which r — 2'v4, say 6 . Inereiore, a = 

7T X 12000 ' ' J 

6"; and since the assumed length = 6", the journal area = 36 sq. in. 

Then if the allowable pressure per square inch = 700 lbs., the total 

allowable working pressure --- 25200 lbs. This is greater than the 

possible working pressure, and hence the lubricant would not be 

squeezed out. The size of both journal and crank-pin is therefore 

dictated by the maximum bending moment. 

104. To design the cross-head pin for the same engine. — In Fig. 

110, C represents the cross-head pin. The force P, = 80000, may 

be applied as indicated. The pin is supported at both ends, and 

the connecting-rod box bears upon it throughout its entire length, 

AD or BE. The pin would yield by shearing on the sections AB 

and DE. The shearing strength of the machinery steel, of which it 

would be made, may be assumed equal to 50000 lbs. A stress of 

8000 lbs. would therefore give a factor of safety of 6 -f-. The neces- 

. . ., ,80000 .. . . 

sary area in shear would equal = 10, or 5 square inches for 

each section. This corresponds to a diameter of 2'5 ~f . The length 
of pin may be found as follows : Find the mean working force upon 
the pin, by drawing the indicator card, as modified by acceleration 
of reciprocating parts, and multiplying the value of its mean 
ordinate, in pounds per square inch, by the piston area. The value 
for this case = about 12000 lbs. The allowable pressure per square 
inch of journal = 800 lbs. Hence the journal area = 12000 -s- 800 
= 15. The length then — 15 h- 25 = 6". The diameter of the 
cross-head pin, therefore, is dictated by the applied force, while its 
length depends upon the maintenance of lubrication. The judg- 
ment of the designer might require this pin to be still larger to re- 
duce wear and to maintain its form. 

Journals whose maintenance of form is of chief importance, 
must be designed from precedent, or according to the judgment of 
the designer. No theory can lead to correct proportions. In fact 



DESIGN OF JOURNALS. 117 

these proportions are eventually determined by the process of 
Machine Evolution. 

105. Thrust Journals. — When a rotating machine part is sub- 
jected to pressure parallel to the axis of rotation, means must be 
provided for the safe resistance of that pressure. In the case of 
vertical shafts the pressure is due to the weight of the shaft and its 
attached parts — as the shafts of turbine water-wheels that rotate 
about vertical axes. In other cases the pressure is due to the work- 
ing force — as the shafts of propeller wheels, the spindles of a chuck- 
ing lathe, etc. The end thrust of vertical shafts is very often 
resisted by the "squared up" end of the shaft. This is in- 
serted in a bronze or brass " bush," which embraces it to 
prevent lateral motion, as in Fig. 111. If the pressure be too 
great, the end of the shaft may be enlarged so as to increase the 
bearing surface, thereby reducing the pressure per square inch. 
This enlargement must be within narrow limits, however. See Fig. 
112. AB is the axis of rotation, and ACD is the rotating part, its 
bearing being enlarged at CD. Let the conditions of wear be con- 
sidered. The velocity of rubbing surface varies from zero at the 
axis to a maximum at C and D. It has been seen that the increase 
of the velocity of rubbing surface increases both the force of the 
friction and the space through which that force acts ; it therefore 
increases the work of the friction, and therefore the tendency to 
wear. From this it will be seen that the tendency to wear increases 
from the centre to the circumference of this " radial bearing," and 
that, after the bearing has run for a while, the pressure will be 
localized near the centre, and heating and abrasion may result. 
For this reason, where there is severe stress to be resisted, the bear- 
ing is usually divided up into several parts, the result being what 
is known as a "collar thrust bearing," as shown in Fig. 113. By 
the increase in the number of collars, the bearing surface may be 
increased without increasing the tendency to unequal wear. The 
radial dimension of the bearing is kept as small as is consistent 
with the other considerations of the design. It is found that the 
"tractrix," the curve of constant tangent, gives the same work of 



118 MACHINE DESIGN. 

friction, and hence the same tendency to wear in the direction of 
the axis of rotation, for all parts of the wearing surface. (See 
"Church's Mechanics," page 181.) This is without doubt the best 
form for a thrust bearing, but the difficulties in the way of the 
accurate production of its curved outlines have interfered with its 
extensive use. 

The pressure that is allowable per square inch of projected area 
of the bearing surface varies in thrust bearings with several con- 
ditions, as it does in journals subjected to pressure at right angles 
to the axis. Thus in the pivots of turn-tables, swing bridges, 
cranes, and the like, the movement is slow and never continuous, 
often being reversed; and also the conditions are such that "bath 
lubrication" may be used, and the allowable unit pressure is very 
high — equal often to. 1500 pounds per square inch, and in some 
cases greatly exceeding that value. The following table may be 
used as an approximate guide in the designing of thrust bearings. 
The material of the thrust journal is wrought iron or steel, and 
the bearing is of bronze or brass (babbitt metal is seldom used for 
this purpose). Bath lubrication is used, i. e., the running surfaces 
are submerged constantly in a bath of oil. 

Mean Velocity Allowable Unit Pressure, 
of rubbing lbs. per square inch 

surface, feet of projected area of 

per minute. the rubbing surface. 

Up to 50 1000 

50 to 100 600 

100 to 150 350 

150 to 200 100 

Above 200 50 

If the journal is of cast iron and runs on bronze or brass, the 
values of allowable pressure given should be divided by 2. 

106. Examples to illustrate the design of thrust journals. 

Example I. — It is required to design a thrust journal whose out- 
line is a tractrix. It is required to support a vertical shaft which with 
its attached parts weighs 2000 lbs., and runs at a rotative speed of 200 
revolutions per minute. The dimensions of the thrust journal are 



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DESIGN OF JOURNALS. 119 

as yet unknown, and therefore the velocity of rubbing surface must 
be estimated. Suppose that the mean diameter of the journal is 2"; 
then the mean velocity of rubbing surface will be2X~XiV-f-12 
= 103 feet per minute. This is so near the limit in the table 
between an allowable pressure of 350 and 600 that an intermediate 
value may be used, say 450 pounds. The projected area of the 
journal then will equal the total pressure divided by the allowable 
pressure per square inch of the journal =2000 -f- 450 = 444 square 
inches. The journal must not be pointed, as in (a) Fig. 114, but 
must be as shown in (b). The dimension BC may be assumed 
equal to 1". The projected area of the journal is equal to the cir- 
cular area whose diameter is AD, minus the circular area whose 
diameter is BC, and this may be equated with the required value, 
equal 4"44, and the equation solved for the required dimension, AD. 

{ADy* (BC)** AA 
Thus - — t 2 — t— = 4-44 



4-44. v 4 
Therefore {AD)* = + (BC)*. 



4D=. 1 /6-j68 = 2-58'\ 

In order now to draw the required journal, lay off from the axis 
EF the distance EG, equal half AD, and through the point G 
draw a tractrix whose constant tangent is equal to EG, con- 
tinuing the curve till it reaches a point C, such that FC is equal to 
half the assumed value of BC. The vertical dimension of the jour- 
nal is thereby determined, and the corresponding curve, BH, may 
be drawn on the other side of the axis EF. 

107. Example II. — It is required to design the collar thrust jour- 
nal that is to receive the propelling pressure from the screw of a small 
yacht. The necessary data are as follows : The maximum power 
delivered to the shaft is 70 H. P.; pitch of screw is 4 feet ; slip of 
screw is 20,% ; shaft revolves 250 times per minute ; diameter of 
shaft is 4" . 

For everv revolution of the screw the yacht moves forward a 



120 MACHINE DESIGN. 

distance = 4 ft. less 20% = 3*2 ft., and the speed of the yacht in feet 
per minute = 250 X 32 — 800. 70 H. P. = 70 X 33000 = 2,310,000 
foot-pounds per minute. This work may be resolved into its factors 
of force and space, and the propelling force is equal to 2,310,000 -*- 
800 = 2900 lbs. nearly. The shaft is 4" diameter, and the collars 
must project beyond its surface. Estimate that the mean radius of 
the rubbing surface is 4*5", then the mean velocity of rubbing sur- 
face would equal 4-5X^-^-12X 250=294 feet per minute. The 
allowable value of pressure per square inch of journal surface for a 
velocity above 200 ft. per minute is 50 lbs. The necessary area of 
the journal surface is therefore = 2900 -r- 50 = 58 square inches. 
It has been seen that it is desirable to keep the radial dimension of 
the collar surface as small as possible in order to have as nearly 
the same velocity at all parts of the rubbing surface as possible. 
The width of collar in this case will be assumed = 0'75"; then the 
bearing surface in each collar 

= 5 ' 5 ' * * - ^^ = 237 - 12-5 = 11-2. 
4 4 

Then the number of collars equals the total required area divided by 
the area of each collar = 58 -r- 11*2 = 5'18, say 6. 

108. Bearings and Boxes. — The function of a bearing or box is 
to insure that the journal with which it engages shall have an 
accurate motion of rotation or vibration about the given axis. It 
must therefore fit the journal without lost motion ; must afford 
means of taking up the lost motion that results necessarily from 
wear ; must resist the forces that come upon it through the jour- 
nal, without undue yielding ; must have the wearing surface of such 
material as will run in contact with the material of the journal 
with the least possible friction, and least tendency to heating and 
abrasion ; and must usually include some device for the main- 
tenance of the lubrication. The selection of the materials and the 
providing of sufficient strength and stiffness depends upon prin- 
ciples already considered, and so it remains to discuss the means 
for the taking up of necessary wear and for providing lubrication. 



DESIGN OF JOURNALS. 121 

Boxes are sometimes made solid rings or shells, the journal being 
inserted endwise. In this case the wear can only be taken up by 
making the engaging surfaces of the box and journal conical, and 
providing for endwise adjustment either of the box itself or of the 
part carrying the journal. Thus, in Fig. 115, the collars for the 
preventing of end motion while running, are jamb nuts, and loose- 
ness between the journal and box may be taken up by moving the 
journal axially toward the left. 

By far the greater number of boxes, however, are made in sec- 
tions, and the lost motion is taken up by moving one or more 
sections toward the axis of rotation. The tendency to wear is 
usually in one direction, and it is sufficient to divide the box 
into halves. Thus, in Fig. 116, the journal rotates about the 
axis 0, and all the wear is due to the pressure P acting in the 
direction shown. The wear will therefore be at the bottom of 
the box. It will suffice for the taking up of wear to dress off the 
surfaces at aa, and thus the box cap may be drawn further down 
by the bolts, and the lost motion is reduced to an admissible value. 
" Liners" or "shims," which are thin pieces of sheet metal, may be 
inserted between the surfaces of division of the box at aa, and may 
be removed successively for the lowering of the box cap as the wear 
renders it necessary. If the axis of the journal must be kept in a 
constant position, the lower half of the box must be capable of being 
raised. 

Sometimes, as in the case of the box for the main journal of 
a steam engine shaft, the direction of wear is not constant. Thus, 
in Fig. 117, A represents the main shaft of an engine. There is a 
tendency to wear in the direction B because of the weight of the 
shaft and its attached parts ; there is also a tendency to wear 
because of the pressure that comes through the connecting-rod and 
crank. The direction of this pressure is constantly varying, but 
the average direction on forward and return stroke may be repre- 
sented by C and D. Provision needs to be made, therefore, for 
taking up wear in these two directions. If the box be divided on 
the line EF, wear will be taken up vertically and horizontally by 

16 



122 MACHINE DESIGN. 

reducing the liners. Usually, however, in the larger engines the 
box is divided into four sections, A, B, C, and D (Fig.. 118), and A 
and C are capable of being moved toward the shaft by means of 
screws or wedges, while D may be raised by the insertion of 
"shims." 

The lost motion between a journal and its box is sometimes 
taken up by making the box as shown in Fig. 11.9. The external 
surface of the box is conical and fits in a conical hole in the 
machine frame. The box is split entirely through at J, parallel to 
the axis, and partly through at B and C. The ends of the box are 
threaded, and the nuts E and F are screwed on. After the journal 
has run long enough so that there is an unallowable amount of lost 
motion, the nut F is loosened and E is screwed up ; the effect being 
to draw the conical box further into the conical hole in the machine 
frame ; the hole through the box is thereby closed up, and 
lost motion is reduced. After this operation the hole cannot be 
truly cylindrical, and if the cylindrical form of the journal has 
been maintained, it will not have a bearing throughout its entire 
surface. This is not usually of very great importance, however, 
and the form of box has the advantage that it holds the axis of the 
journal in a Constant position. 

All boxes in self-contained machines, like engines or machine 
tools, need to be rigidly supported to prevent the localization of 
pressure, since the parts that carry the journals are made as rigid 
as possible. In line shafts and other parts carrying journals, when 
the length is great in comparison to the lateral dimensions, some 
yielding must necessarily occur, and if the boxes were rigid, local- 
ization of pressure would result. Hence "self-adjusting boxes" 
are used. A point in the axis of rotation at the centre of the 
length of the box is held immovable, but the box is free to move 
in any way about this point, and thus adjusts itself to any yield- 
ing of the shaft. This result is attained as shown in Fig. 120. 
is the centre of the motion of the box ; B and A are spherical sur- 
faces formed on the box, their centre being at 0. The support for the 
box carries internal spherical surfaces which engage with A and B. 





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DESIGN OF JOURNALS. 123 

Thus the point is always held in a constant position, but the box 
itself is free to move in any way about as a centre. Therefore 
the box adjusts itself, within limits, to any position of the shaft, 
and hence the localization of pressure is impossible. 

In thrust bearings for vertical shafts the weight of the shaft and 
its attached parts serves to hold the rubbing surfaces in contact, 
and the lost motion is taken up by the shaft following down as wear 
occurs. In collar thrust bearings for horizontal shafts the design 
is such that the bearing for each collar is separate and- adjustable. 
The pressure on the different collars may thus be equalized.* 

109. Lubrication of Journals. — The best method of lubrication is 
that in which the rubbing surfaces are constantly submerged in a 
bath of lubricating fluid. This method should be employed when- 
ever possible, if the pressure and surface velocity are high. Unfor- 
tunately it cannot be used in the majority of cases. Let J, Fig. 
121, represent a journal with its box, and let A, B, and C be oil 
holes. If oil be introduced into the hole A, it will tend to flow out 
from between the rubbing surfaces by the shortest way ; i. <?., it will 
come out at D. A small amount will probably go toward the other 
end of the box because of capillary attraction, but usually none of 
it will reach the middle of the box. If oil be introduced at 0, it 
will come out at E. A constant feed, therefore, might be main- 
tained at A and C, and yet the middle of the box might run dry. 
If the oil be introduced at B, however, it tends to flow equally in 
both directions, and the entire journal is lubricated. The con- 
clusion follows that oil ought, when possible, to be introduced at 
the middle of the length of a cylindrical journal. If a conical 
journal runs at a high velocity, the oil under the influence of cen- 
trifugal force tends to go to the large end of the cone, and therefore 
the oil should be introduced at the small end to insure its distribu- 
tion over the entire journal surface. 

If the end of a vertical thrust journal, whose outline is a cone or 
tractrix, as in Fig. 122, dips into a bath of oil, B, the oil will be 

*For complete and varied details of marine thrust bearings see " Maw's 
Modern Practice in Marine Engineering." 



124 MACHINE DESIGN. 

carried by its centrifugal force, if the velocity be high, up between 
the rubbing surfaces, and will be delivered into the groove AA. If 
holes connect A and 7?, gravity will return the oil to B, and a con- 
stant circulation will be maintained. If the thrust journal has 
simply a flat end, as in Fig. 123, the oil should be supplied at the 
centre of the bearing; centrifugal force will then distribute it over 
the entire surface. Vertical shaft thrust journals may usually be 
arranged to run in an oil bath. Marine collar thrust journals are 
always arranged to run in an oil bath. 

Sometimes a journal is stationary and the box rotates about it, 
as in the case of a loose pulley, Fig. 124. If the oil is introduced 
into a tube A, as is often done, its centrifugal force will carry it 
away from the rubbing surface. But if a hole is drilled in the axis 
of the journal, the lubricant introduced into it will be carried to the 
rubbing surfaces as required. If a journal is carried in a rotating 
part at a considerable distance from the axis of rotation, and it 
requires to be oiled while in motion, a channel may be provided 
from the axis of rotation where oil may be introduced conveniently, 
to the rubbing surfaces, and the oil will be carried out by centrifu- 
gal force. Thus Fig. 125 shows an engine crank in section. Oil 
is introduced at 0, and centrifugal force carries it through the 
channel provided to a, where it serves to lubricate the rubbing sur- 
faces of the crank-pin and its box. If a journal is carried in a 
reciprocating machine part, and requires to be oiled while in 
motion, the a wick and wiper" method is one of the best. See Fig. 
126. An ordinary oil cup with an adjustable feed is mounted in a 
proper position opposite the end of the stroke of the reciprocating 
part, and a piece of flat wick projects from its delivery tube. A 
drop of oil runs down and hangs suspended at its end. Another 
oil cup is attached to the reciprocpting part, which carries a hooked 
" wiper," B. The delivery tube from C leads to the rubbing surfaces 
to be lubricated. When the reciprocating part reaches the end of 
its stroke the wiper picks off the drop of oil from the wick, and it 
runs down into the oil cup 0, and thence to the surfaces to be 
lubricated. This method applies to the oiling of the cross-head pin 



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DESIGN OF JOURNALS. 125 

of a steam engine. The same method is sometimes applied to the 
crank-pin, but here, through a part of the revolution, the tendency 
of the centrifugal force is to force the oil out of the cup, and there- 
fore the plan of oiling from the axis is probably preferable. 

When journals are lubricated by feed oilers, and are so located 
as not to attract attention if the lubrication should fail for any 
reason, " tallow boxes " are used. These are cup-like depressions 
usualty cast in the box cap, and communicating by means of an 
oil hole with the rubbing surface. These cups are filled with grease 
that is solid at the ordinary temperature of the box, but if there 
is the least rise of temperature because of the failure of the oil 
supply, the grease melts and runs to the rubbing surfaces, and sup- 
plies the lubrication temporarily. This safety device is used very 
commonly on line shaft journals. 

The most common forms of feed oilers are : I. The oil cup with 
an adjustable valve that controls the rate of flow. II. The oil 
cup with a wick feed, Fig. 127. The delivery has a tube in- 
serted in it which projects nearly to the top of the cup. In this 
tube a piece of wicking is inserted, and its end dips into the oil in 
the cup. The wick, by capillary attraction, carries the oil slowly 
and continuously over through the tube to the rubbing surfaces. 
III. The cup with a copper rod, Fig. 128. The oil cup is filled 
with grease that melts with a very slight elevation of temperature, 
and A is a small copper rod dropped into the delivery tube and 
resting on the surface of the journal. The slight friction between 
the rod and the journal warms the rod and it melts the grease 
in contact with it, which runs down the rod to the rubbing sur- 
face. IV. Sometimes a part of the surface of the bottom half of 
the box is cut away and a felt pad is inserted, its bottom being in 
contact with an oil bath. This pad rubs against the surface of the 
journal, is kept constantly soaked with oil, and maintains lubrication. 



CHAPTER XI. 



SLIDING SURFACES. 



110. So much of the accuracy of action of machines depends on 
the sliding surfaces that their design deserves the most careful 
attention. The perfection of the cross-sectional outline of the cylin- 
drical or conical forms produced in the lathe, depends on the per- 
fection of form of the spindle. But the perfection of the outlines of 
a section through the axis depends on the accuracy of the sliding 
surfaces. All of the surfaces produced by planers, and most of 
those produced by milling machines, are dependent for accuracy on 
the sliding surfaces in the machine. 

Suppose that the short block A, Fig. 129, is the slider of a slider- 
crank chain, and that it slides on a relatively long guide, D. The 
direction of rotation of the crank, a, is as indicated by the arrow. 
B and C are the extreme positions of the slider. The pressure 
between the slider and the guide is greatest at the mid-position, A, 
and at the extreme positions, B and C, it is only the pressure due 
to the weight of the slider. Also the velocity is a maximum when 
the slider is in its mid-position, and decreases toward the ends, 
becoming zero when the crank a is on its centre. The work of fric- 
tion is therefore greatest at the middle, and is very small near the 
ends. Therefore the wear would be greatest at the middle, and the 
guide would wear concave. If now the accuracy of a machine's 
working depends on the perfection of A's rectilinear motion, that 
accuracy will be destroyed as the guide D wears. Suppose a gib, 
EFG, to be attached to A, Fig. 130, and to engage with D, as shown, 
to prevent vertical looseness between A and D. If this gib be taken 
up to compensate wear after it has occurred, it will be loose in the 



SLIDING SURFACES. 127 

middle position when it is tight at the ends, because of the unequal 
wear. Suppose that A and D are made of equal length, as in Fig. 
131. Then when A is in the mid-position corresponding to maxi- 
mum pressure, velocity, and wear, it is in contact with D through- 
out its entire surface, and the wear is therefore the same in all parts 
of the surface. The slider retains its accuracy of rectilinear motion 
regardless of the amount of wear, the gib may be set up, and will 
be equally tight in all positions. 

HA and B, Fig. 132, are the extreme positions of a slider, D 
being the guide, a shoulder would be finally worn at C. It would 
be better to cut away the material of the guide, as shown by the 
dotted line. Slides should always " wipe over " the ends of the 
guide when it is possible. Sometimes it is necessary to vary the 
length of stroke of a slider, and also to change its position relatively 
to the guide. Examples: " Cutter bars " of slotting and shaping 
machines. In some of these positions, therefore, there will be a 
tendency to wear shoulders in the guide and also in the cutter bar 
itself. This difficulty is overcome if the slide and guide are made 
of equal length, and the design is such that when it is necessary to 
change the position of the cutter bar that is attached to the slide, 
the position of the guide may be also changed so that the relative 
position of slide and guide remains the same. The slider surface 
will then just completely cover the surface of the guide in the mid- 
position, and the slider will wipe over each end of the guide, what- 
ever the length of the stroke. 

In many cases it is impossible to make the slider and guide of 
equal length. Thus a lathe carriage cannot be as long as the bed ; 
a planer table cannot be as long as the planer bed, nor a planer 
saddle as long as the cross-head. When these conditions exist 
especial care should be given to the following : I. The bearing 
surface should be made so large in proportion to the pressure to be 
sustained that the maintenance of lubrication shall be insured 
under all conditions. II. The parts which carry the wearing sur- 
faces should be made so rigid that there shall be no possibility of 
the localization of pressure from yielding. 



128 MACHINE DESIGN. 

111. As to form, guides may be divided into two classes : angu- 
lar guides and flat guides. Fig. 133 a, shows an angular guide, 
the pressure being applied as shown. The advantage of this form 
is, that as the rubbing surfaces wear, the slide follows down and 
takes up both the vertical and lateral wear. The objection to 
this form is that the pressure is not applied at right angles to 
the wearing surfaces, as it is in the flat guide shown in b. But 
in b a gib, A, must be provided to take up the lateral wear. The 
gib is either a wedge or a strip with parallel sides backed up by 
screws. Guides of these forms are used for planer tables. The 
weight of the table itself holds the surfaces in contact, and if the 
table is light the tendency of a heavy side cut would be to force the 
table up one of the angular surfaces away from the other. If the 
table is very heavy, however, there is little danger of this, and 
hence the angular guides of large planers are much flatter than 
those of smaller ones. In some cases one of the guides of a planer 
table is angular and the other is flat. The side bearings of the flat 
guide may then be omitted, as the lateral wear is taken up by the 
angular guide. This arrangement is undoubtedly good if both 
guides wear down equally fast. 

112. Fig. 134 shows three forms of sliding surfaces such as are 
used for the cross slide of lathes, the vertical slide of shapers, the 
table slide of milling machines, etc. A is a taper gib that is forced 
in by a screw at D to take up wear. When it is necessary to take 
up wear at B, the screw may be loosened and a shim or liner may 
be inserted between the surfaces at a. C is a thin gib, and the wear 
is taken up by means of several screws like the one shown. This 
form is not so satisfactory as the wedge gib, as the bearing is chiefly 
under the points of the screws, the gib being thin and yielding, 
whereas in the wedge there is complete contact between the metallic 
surfaces. 

113. The sliding surfaces thus far considered have to be designed 
so that there will be no lost motion while they are moving, because 
they are required to move while the machine is in operation. The 
gibs have to be carefully designed and accurately set so that the 




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SLIDING SURFACES. 129 

moving part shall be just " tight and loose," i. e., so that it shall be 

free to move, without lost motion to interfere with the accurate 

action of the machine. There is, however, another class of sliding 

parts, like the sliding head of a drill press, or the tail stock of a 

lathe, that are never required to move while the machine is in 

operation. It is only required that they shall be capable of being 

fastened accurately in a required position, their movement being 

simply to readjust them to other conditions of work, while the 

machine is at rest. No gib is necessary and no accuracy of motion 

is required. It is simply necessary to insure that their position is 

accurate when they are clamped for the special work to be done. 
17 



CHAPTER XII. 

BOLTS AND SCREWS AS MACHINE FASTENINGS. 

114. Classification may be made as follows : I. Bolts. II. 
Studs. III. Cap Screws, or Tap Bolts. IV. Set Screws. V. Ma- 
chine Screws. 

A "bolt" consists of a head and round body on which a thread 
is cut, and upon which a nut is screwed. When a bolt is used to 
connect machine parts, a hole the size of the body of the bolt is 
drilled entirely through both parts, the bolt is put through, and the 
nut screwed down upon the washer. See Fig. 135. 

A " stud " is a piece of round metal with a thread cut upon each 
end. One end is screwed into a tapped hole in some part of a 
machine, and the piece to be held against it, having a hole the size 
of the body of the stud, is put on, and a nut is screwed upon the 
other end of the stud against the piece to be held. See Fig. 136. 

A "cap screw" is a substitute for a stud, and consists of a head 
and body on which a thread is cut. See Fig. 137. The screw is 
passed through the removable part and screwed into a tapped hole 
in the part to which it is attached. A cap screw is a stud with a 
head substituted for the nut. 

A hole should never be tapped into a cast iron machine part 
when it can be avoided. Cast iron is not good material for the 
thread of a nut, since it is weak and brittle and tends to crumble. 
In very many cases, however, it is absolutely necessary to tap into 
cast iron. It is then better to use studs, if the attached part needs to 
be removed often, because studs are put in once for all, and the cast 
iron thread would be worn out eventually if cap screws were used. 

When one machine part surrounds another, as a pulley hub 



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BOLTS AND SCREWS AS MACHINE FASTENINGS. 131 

surrounds a shaft, relative motion of the two is often prevented by 
means of a "set screw" which is a threaded body with a small 
square head. Fig. 138. The end is either pointed as in Fig. 138 b, 
or cupped as in c, and is forced against the inner part by screwing 
through a tapped hole in the outer part. 

The term " machine screws " covers many forms of small screws, 
usually with screw-driver heads. All of the kinds given in this 
classification are made in great variety of size, form, length, etc. 

115. Design of Bolts and Screws. — Tensile stress is induced in a 
bolt by tightening it. This stress may equal, or very greatly exceed, 
the tensile stress due to working forces. The stress due to tightening 
may be approximately found as follows : In Fig. 139, a force Pis 
applied to the wrench handle at A. During the turning of the 
wrench the ratio of movement of the point A to the movement of 
the nut axially = 2*1 to p, in which I is the wrench lever arm, and 
p is the pitch of the screw (axial distance between threads). If 
there were no frictional resistance, the force P and the resistance to 
tension of the bolt, == T, would be in equilibrium at the instant of 
tightening up, and the following equation would be true: 

P.2*l=Tp.., 

But force P must also overcome the frictional resistance between 
the nut and the screw thread, and between the nut and washer. 
This resistance R = Tf ; in which /= the coefficient of friction. 
The radius, r, of this resistance, R, may be assumed, for this ap- 
proximation, equal to the radius of the top of the screw threads X 
1*5. The ratio of movement of R to axial movement of the nut = 
2*r to p. At the instant of tightening up there is equilibrium 
between P, R, and T, and the following equation is true: 

2 7tpl=Tp J r 2nR r ; substituting for R =fT, 

= Tp + 2nTfr, 

, 2nPl 

whence 1 = — , n . . 

p + 2*/r 



132 MACHINE DESIGN. 

For a i" bolt the values are I = 8", p = 0*077",/= 0*15, r = 0*875. 
Making P-~- 1 lb., 7"= 116 lbs. Hence, for every pound applied at 
A there results 116 lbs. tensile stress in the bolt. The ultimate 
strength of the bolt -j- 116 equals the force applied at A necessary 
to break the bolt. The area of cross-section at the bottom of the 
thread of a \" bolt =0*12 sq. in. Assume the ultimate strength of 
the material of the bolt — 50000 lbs. per sq. in. The ultimate ten- 
sile strength of the bolt ■= 50000 X 0*12 = 6000 lbs. Then the force 
at A to break the bolt = 6000 -*- 116 = 52 lbs. nearly. This is prob- 
ably not the actual force that would break the bolt, because the 
assumptions are probably somewhat inaccurate ; but it indicates 
that a half inch bolt may be pulled in two by force applied by a 
man to the wrench handle. For a %" bolt the force becomes about 
100 lbs. 

Suppose a nut screwed up with a resulting tensile stress in the 
bolt = p. Suppose that a gradually increasing working force, = p xi 
is applied. If there were no elongation of the bolt, the total stress 
in the bolt would equal p -f- p v But elongation does result from 
the application of p v and p is reduced, and the total stress in the 
bolt is less than p + p r 

116. Illustration. — In Fig. 140 the tensile stress in the bolt due 
to screwing up = p. The pressure between the surfaces in contact 
at CD is therefore — £>. Suppose a working force, p Xi applied tend- 
ing to separate A and B. The bolt yields to the increased stress, 
and the pressure at CD is reduced. The tensile stress in the bolt is 
now equal to the working force plus the reduced pressure at CD. 
When the working force reduces the pressure at CD to zero, the 
stress in the bolt = the working force, and if CD were a steam 
joint, it would leak. 

117. It is required to design the fastenings to hold on the steam 
chest cover of a steam engine. The opening to be covered is rec- 
tangular, 10"x 12". The maximum steam pressure is 100 lbs. per 
square inch. The joint must be held steam tight. Short unyield- 
ing fastenings are therefore best suited to the purpose, and studs 
will be used. They will be made of machinery steel of 60000 lbs. 



BOLTS AND SCREWS AS MACHINE FASTENINGS. 133 

tensile strength, and will be j" outside diameter because smaller 
studs may be ruptured by the force applied in screwing up. It will 
be assumed that the stress on the studs is equal to p -f- p„ i. e., the 
stress due to screwing up, plus the working stress. This assump- 
tion cannot be exactly true, as is seen from the preceding illustra- 
tion ; but the resulting error is on the safe side. The diameter of a 
|" stud at the bottom of the thread is 0"62", and the area is = 
0*62 2 X ~ -s- 4 = 0'3 sq. in. The ultimate strength of the stud is, 
therefore, 03 X 60000 = 18000. The factor of safety may be 4 
because the stress member is of resilient material and is not subject 
to shocks. Then the allowable stress on each stud would be equal 
to 18000 -T- 4 == 4500. The stress due to screwing up, subtracted 
from the total allowable stress gives the allowable working stress in 
the stud. The stress due to screwing up may be found from the 
above equation for T. Assuming P= 30 lbs.; £ = 10"; f=0'lo; 
r — 0'56; gives T= 3000 lbs. nearly. The allowable working stress 
in each stud equals 4500 — 3000 = 1500 lbs. The maximum work- 
ing force on the cover equals the area of the opening covered, multi- 
plied by the maximum working pressure per square inch; = 
10 X 12 = 120 sq. in. X 100 lbs. per sq. in. =±= 12000 lbs. This di- 
vided by the allowable working pressure for each stud gives the 
number of studs required for strength, = 12000 -s- 1500 = 8. There- 
fore 8 studs will serve for strength. But in order to make a steam 
tight joint, with a reasonable thickness of steam chest cover, the 
distance between the stud centres should not be greater than about 
4-J". The opening is 10"xl2", as shown in Fig. 141. There must be 
a band about j" wide around this for making the joint upon which 
the studs must not encroach. This makes the distance between 
the vertical rows of studs 14", and between the horizontal rows 12". 
The whole length over which the studs are to be distributed then — 
12 -\- 12 -f- 14 -f 14 = 52". If they are 4 , 5" apart the number of 
studs = 52 -i- 45 = 11*5. Hence it is necessary to use 12 studs to 
make the joint tight, while 8 would serve for strength. In order 
to get a symmetrical arrangement, it will probably be necessary to 
use 14 studs. The number of studs is therefore dictated by the 



134 MACHINE DESIGN. 

conditions necessary to maintain a tight steam joint, and not by 
the applied forces. 

118. The elongation of a bolt with a given total stress, depends 
upon the length and area of its least cross-section. Suppose, to 
illustrate, that the bolt, Fig. 142, has a reduced section over a 
length I as shown. This portion A has less cross-sectional area 
than the rest of the bolt, and when any tensile force is applied, the 
resulting unit stress will be greater in A than elsewhere. The unit 
strain, or elongation, will be proportionately greater up to the 
elastic limit ; and if the elastic limit is exceeded in the portion A, 
the elongation there will be, far greater than elsewhere. If there is 
much difference of area and the bolt is tested to rupture, the elonga- 
tion will be chiefly at A. There would be a certain elongation per 
inch of A at rupture. Hence, the greater the length of A, the 
greater the total elongation of the bolt. If the bolt had not been 
reduced at A, the minimum section would be at the root of the 
screw threads. The axial length of this section is very small. 
Hence the elongation at rupture would be small. Suppose there 
are two bolts, A with, and B without, the reduced section. They 
are alike in other respects. They are subjected to equal tensile 
shocks. Let the energy of the shock == E. This energy is divided 
into force and space factors by the resistance of the bolts. The 
space factor equals the elongation of the bolt. This is greater 
in A than in B because of the yielding of the reduced section. 
But the product of force and space factors is the same in both 
bolts, = E; hence the resulting stress in the minimum section 
is less for A than for B. The stress in A may be less than the 
breaking stress; while the greater stress in B may break it. 
The capacity of the bolt to, resist shock is therefore increased by 
lengthening its minimum section to increase the yielding and reduce 
stress. This is not only true of bolts, but of all stress members in 
machines. 

The whole body of the bolt might have been reduced, as shown 
by the dotted lines in Fig. 142, with resulting increase of capacity 
to resist shock. Turning down a bolt, however, weakens it to 




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BOLTS AND SCREWS AS MACHINE FASTENINGS. 135 

resist torsion and flexure, because it takes off the material which 
is most effective in producing large polar and rectangular moments 
of inertia of cross-section. If the cross-sectional area is reduced by 
drilling a hole as shown in Fig. 143, the torsional and transverse 
strength is but slightly decreased, but the elongation will be as 
great, with the same area, as if the area had been reduced by turn- 
ing down. 

119. Prof. Sweet had a set of bolts prepared for special test. 
The bolts were 1 J" -diameter and about 12" long. They were made 
of high grade wrought iron, and were duplicates of the bolts used at 
the crank end of the connecting-rods of one of the standard sizes of 
the Straight Line Engine. Half of the bolts were left solid, while 
the other half were carefully drilled to give them uniform cross- 
sectional area throughout. The tests were made under the direction 
of Prof. Carpenter at the Sibley College Laboratory. One pair of 
bolts was tested to rupture by tensile force gradually applied. The 
undrilled bolt broke in the thread with a total elongation of 0*25". 
The drilled bolt broke between the thread and the bolt head with a 
total elongation of 2 , 25". If it be assumed that the mean force ap- 
plied was the same in both cases, it follows that the total resilience 
of the drilled bolt was nine times as great as that of the solid one. 
" Drop tests," i. e., tests which brought tensile shock to bear upon 
the bolts, were made on other similar pairs of bolts, which tended to 
confirm the general conclusion. 

120. It is required to design proper fastenings for holding on 
the cap of a connecting-rod like that shown in Fig. 144. These 
fastenings are required to sustain shocks, and may be subjected to 
a maximum accidental stress of 20000 lbs. There are two fasten- 
ings, and therefore each must be capable of sustaining safely a 
stress of 10000 lbs. They should be designed to yield as much as 
is consistent with strength ; in other words, they should be tensile 
springs to cushion shocks and thereby reduce the resulting force 
they have to sustain. Bolts should therefore be used, and the 
weakest section should be made as long as possible. Wrought iron 
will be used whose tensile strength is 50000 lbs. per square inch. 



136 MACHINE DESIGN. 

The stress given is the maximum accidental stress, and is four 
times the working stress. It is not, therefore, necessary to give the 
bolts great excess of strength over that necessary to resist actual 
rupture by the accidental force. Let the factor of safety be 2. 
Then the cross-sectional area of each bolt must be such that it will 
just sustain 10000 X 2 = 20000 lbs. This area = 20000 -=- 50000 
= 0*4 square inches. This area corresponds to a diameter of 0"71", 
and that is nearly the diameter of a -J" bolt at the bottom of the 
thread ; hence -J" bolts will be used. The cross-sectional area of 
the body of the bolt must now be made at least as small as that at 
the bottom of the thread. This may be accomplished by drilling. 

121. When bolts are subjected to constant vibration there is a 
tendency for the nuts to loosen. There are many ways to prevent 
this, but the most common one is by the use of jamb nuts. Two 
nuts are screwed on the bolt ; the under one is set up against the 
surface of the part to be held in place, and then while this nut is 
held with a wrench the other nut is screwed up against it tightly. 
Suppose that the bolt has its axis vertical and that the nuts are 
screwed on the upper end. The nuts being screwed against each 
other the upper one has its internal screw surfaces forced against 
the under screw surfaces of the bolt, and if there is any lost motion, 
as there almost always is, there will be no contact between the 
upper surfaces of the screw on the bolt and the threads of the nut. 
Just the reverse is true of the under nut ; i. e., there is no contact 
between the under surfaces of the threads on the bolt and the 
threads on the nut. Therefore no pressure that comes from the 
under side of the under nut can be communicated to the bolt 
through the under nut directly, but it must be received by the 
upper nut and communicated by it to the bolt, since it is the upper 
nut alone that has contact with the under surfaces of the thread. 
Therefore the jamb nut, which is usually made about half as thick 
as the other, should always be put on next to the surface of the 
piece to be held in place. 



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CHAPTER XIII. 

MEANS FOR PREVENTING RELATIVE ROTATION. 

122. Keys are chiefly used to prevent relative rotation between 
shafts and the pulleys, gears, etc. which they support. Keys may 
be divided into parallel keys, taper keys, and feathers or splines. 

For a parallel key the " seat," both in the shaft and the attached 
part, has parallel sides, and the key simply prevents relative ro- 
tary motion. Motion parallel to the axis of the shaft must be pre- 
vented by some other means ; as by set screws which bear upon the 
top surface of the key, as shown in Fig. 145. A parallel key 
should fit accurately on the sides and loosely at the top and bottom . 

A taper key has parallel sides and has its top and bottom sur- 
faces tapered, and is made to fit on all four surfaces, being driven 
tightly " home." It prevents relative motion of any kind between 
the parts connected. If a key of this kind has a head, as shown in 
Fig. 146, it is called a " draw key," because it is drawn out when 
necessary, by driving a wedge between the hub of the attached part 
and the head of the key. When a taper key has no head it is re- 
moved by driving against the point with a " key drift." 

Feathers or Splines are keys that prevent relative rotation, but 
purposely allow axial motion. They are sometimes made fast in 
the shaft, as in Fig. 147, and there is a key "way " in the attached 
part that slides along the shaft. Sometimes the feather is fast- 
ened in the hub of the attached part, as shown in Fig. 148, and 
slides in a long key way in the shaft. 

John Richards' rule for keys is (see Fig. 149) w = -j. t has 

such value that « = 30°. This rule is deviated from somewhat, as 



138 MACHINE DESIGN. 

shown by the following table taken from Richards' " Manual of 
Machine Construction," page 58. 

2i 3 3i 4 5 6 7 8 
I I i 1 1* If H H 

I 7 x6 4 I "16 I3 x6 J 1 

When d exceeds 8" two or more keys should be used, and w may 
then — d -i- 16 ; £ being as before of such value that a shall == 30°. 
The following table for dimensions for parallel keys is also from 
Richards' "Manual": 



w = l li 


H 


If 2 


d=i s l6 


1 


7 x6 i 


« = 5 32 3 x6 


1 

4 


9 32 5 x6 



d=l li 14 


1| 2 2i 3 34 4 


W = 5 32 7 32 9 32 


ii 13 15 17 9 ^ 11 , 

32 ^32 32 '32 '16 16 


t = 3 x6 i 5 x6 


f 7 .6 i «i 1 i 


his for feathers : 




d.= li li If 


2 2| 2| 3 8'f 4 4^ 


w = i i 5 i6 


S ,6 | f i 9 ,6 9 !6 1 


* = 1 1 7 i6 


'■6 4 4 S 1 1 J 



For keying hand wheels and other parts that are not subjected 
to very great stress, a cheap and satisfactory method is to use a 
round key driven into a hole drilled in the joint, as in Fig. 150. . If 
the two parts are of different material, one much harder than the 
other, this method should not be used, as it is almost impossible in 
such case to make the drill follow the joint. 

The taper of keys varies from i" to ¥ to the foot. 

A cotter is a key that is used to attach parts subjected to a force 
of tension tending to separate them. Thus piston rods are often 
connected to both piston and cross-head in this way. Also the sec- 
tions of long pump-rods, etc. 

Fig. 151 shows machine parts held against tension by cotters. 
It is seen that the joint may yield by shearing the cotter at AB 



MEANS FOR PREVENTING RELATIVE ROTATION. 



139 



and CD ; or by shearing CPQ and ARS ; by shearing on the sur- 
faces MO and LN ; or by tensile rupture of the rod on a horizontal 
section at LM. All of these sections should be sufficiently large to 
resist the maximum stress safely. The difficulty is usually to get 
LM strong enough in tension ; but this may usually be accom- 
plished by making the rod larger, or the cotter thinner and wider. 
It is found that taper surfaces if they be smooth and somewhat 
oily will just cease to stick together when the taper equals 1*5" per 
foot. The taper of the rod in Fig. 151 should be about this value 
in order that it may be removed conveniently when necessary. 

123. Shrink and Force Fits. — Relative rotation between machine 
parts is also prevented sometimes by means of shrink and force fits. 
In the former the shaft is made larger than the hole in the part to 
be held upon it, and the metal surrounding the hole is heated, usu- 
ally to low redness, and because of the expansion it may be put on 
the shaft and on cooling, it shrinks and "grips" the shaft. A key 
is sometimes used in addition to this. 

Force fits are made in the same way except that they are put 
together cold, either by driving together with a heavy sledge or by 
forcing together by hydraulic pressure. The necessary allowance 
for forcing, i. e., the excess of shaft diameter over the diameter of 
the hole, is given in the following table : 



Inches. 



Diameter of Shaft 
Allowance for Forcing 



1 
0*004 



2 
0'005 



3 
0-006 



4 
0-006 



5 
0-007 



6 

0-008 



7 
0-008 



8 9 
0-009 0-01 



10 
0-01 



Experience shows that with this allowance a steel shaft may be 
forced into a hole in cast iron by a total pressure of from 40 to 90 
tons. There is no need of keying when parts are put together in 
this way. 



CHAPTER XIV. 

FORM OF MACHINE PARTS AS DICTATED BY STRESS. 

124. Suppose that A and B, Fig. 152, are two surfaces in a 
machine to be joined by a member subjected to simple tension. 
What is the proper form for the member? The stress in all sections 
of the member at right angles to the line of application, A B, of the 
force, will be the same. Therefore the areas of all such sections 
should be equal; hence the outlines of the member should be straight 
lines parallel to AB. The distance of the material from the axis 
AB has no effect on its ability to resist tension. Therefore there is 
nothing in the character of the stress that indicates the form of the 
cross-section of the member. The form most cheaply produced, 
both in the rolling mill and the machine shop, is the cylindrical 
form. Economy, therefore, dictates the circular cross-section. 
After the required area necessary for safely resisting the stress is 
determined, it is only necessary to find the corresponding diameter, 
and it will be the diameter of all sections of the required member 
if they are made circular. Sometimes in order to get a more har- 
monious design, it is necessary to make the tension member just 
considered of rectangular cross-section, and this is allowable al- 
though it almost always costs more. The thin, wide, rectangular 
section should be avoided, however, because of the difficulty of in- 
suring a uniform distribution of stress. A unit stress might result 
from this at one edge, greater than the strength of the material, 
and the piece would yield by tearing, although the average stress 
might not have exceeded a safe value. 

If the stress be compression instead of tension, the same consid- 
erations dictate its form as long as it is a " short block," i. e., as long 




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FORM OF MACHINE PARTS AS DICTATED BY STRESS. 141 

as the ratio of length to lateral dimensions is such that it is sure to 
yield by crushing instead of by "buckling." A short block, there- 
fore, should have its longitudinal outlines parallel to its axis, and i 
its cross-section may be of any form that economy or appearance 
may dictate. Care should be taken, however, that the least lateral 
dimension of the member be not made so small that it is thereby 
converted into a "long column." 

If the ratio of longitudinal to lateral dimensions is such that the 
member becomes a "long column," the conditions that dictate the 
form are changed, because it would yield by buckling or flexure, 
instead of crushing. The strength and stiffness of a long column 
are proportional to the moment of inertia of the cross-section about 
a gravity axis at right angles to the plane in which the flexure oc- 
curs. A long column with "fixed" or "rounded" ends has a tend- 
ency to yield by buckling which is equal in all directions. Therefore 
the moment of inertia needs to be the same about all gravity axes, 
and this of course points to a circular section. Also the moment of 
inertia should be as large as possible for a given weight of material, 
and this points to the hollow section. The disposition of the metal 
in a circular hollow section is the most economical one for long col- 
umn machine members with fixed or rounded ends. This form, 
like that for tension, may be changed to the rectangular hollow sec- 
tion if appearance requires such change. If the long column ma- 
chine member be "pin connected," the tendency to buckle is great- 
est in a plane through the line of direction of the compressive force, 
and at right angles to the axis of the pins. The moment of inertia 
of the cross-section should therefore be greatest about a gravity 
axis parallel to the axis of the pins. Example : a steam engine 
connecting-rod. 

When the machine member is subjected to transverse stress the 
best form of cross-section is probably the I section, a, Fig. 153, in 
which a relatively large moment of inertia, with economy of mater- 
ial, is obtained by putting the excess of the material where it is 
most effective to resist flexure, i. e., at the greatest distance from the 
given gravity axis. Sometimes, however, if the I section has to be 



142 MACHINE DESIGN. 

produced by cutting away the material ate and d, in the machine 
shop, instead of producing the form directly in the rolls, it is 
cheaper to use the solid rectangular section c. If the member sub- 
jected to transverse stress is for any reason made of cast material, 
as is often the case, the form b is preferable, for the following 
reasons: I. The best material is almost sure to be in the thinnest 
part of a casting, and therefore in this case is at / and g, where it 
is most effective to resist flexure. II. The pattern for the form b is 
more cheaply produced and maintained than that for a. III. If 
the surface is left without finishing from the mould, any imperfec- 
tions due to the foundry work are more easily corrected in b than 
in a. Machine members subjected to transverse stress, which con- 
tinually change their position relatively to the force that produces 
the flexure, should have the same moment of inertia about all 
gravity axes. As, for instance, rotating shafts that are strained 
transversely by the force due to the weight of a fly-wheel, or that 
due to the tension of a driving belt. The best form of cross-section 
in this case is circular. The hollow section would give the greatest 
economy of material, but hollow members are expensive to produce 
in wrought material, such as is almost invariably used for shafts. 
Hence the solid circular section is used. 

125. Torsional strength and stiffness are proportional to the polar 
moment of inertia of the cross-section of the member. This is equal 
to the sum of the moments of inertia about two gravity axes at 
right angles to each other. The forms in Fig. 153 are therefore not 
correct forms for the resistance of torsion. The circular solid or 
hollow section, or the rectangular solid or hollow section, should be 
used. 

The I section, Fig. 154, is a correct form for resisting the stress 
P, applied as shown. Suppose the web c to be divided on the 
line CD, and the parts to be moved out so that they occupy the 
positions shown at a and b. The form thus obtained is called a 
" box section." By making this change the moment of inertia 
about AB has not been changed, and therefore the new form is just 
as effective to resist flexure due to the force P as it was before the 



FORM OF MACHINE PARTS AS DICTATED BY STRESS. 143 

change. The box section is better able to resist torsional stress, 
because the change made to convert the I section into the box sec- 
tion has increased the polar moment of inertia. The two forms are 
equally good to resist tensile and compressive force if they are 
sections of short blocks. But if they are both sections of long 
columns, the box section would be preferable, because the moments 
of inertia would be more nearly the same about all gravity axes. 

126. The framing of machines almost always sustains com- 
bined stresses, and if the combination of stresses include torsion, 
flexure in different planes, or long column compression, the box 
section is the best form. In fact the box section is by far the best 
form for the resisting of stress in machine frames. There are 
other reasons, too, beside the resisting of stress that favor its use.* 
I. Its appearance is far finer, giving an idea of completeness that is 
always wanting in the ribbed frames. II. The faces of a box frame 
are always available for the attachment of auxiliary parts without 
interfering with the perfection of the design. III. The strength 
can always be increased by decreasing the size of the core, without 
changing the external appearance of the frame, and therefore with- 
out any work whatever on the pattern itself. The cost of patterns 
for the two forms is probably not very different ; the pattern itself 
being the more expensive in the ribbed form, and the necessary core 
boxes adding to the expense in the case of the box form. The 
expense of production in the foundry, however, is greater for the 
box form than for the ribbed form, because core work is more ex- 
pensive than " green sand " work. The balance of advantage is 
very greatly in favor of box forms, and this is now being recognized 
in the practice of the best designers of machinery. 

127. To illustrate the application of the box form to machine 
members, let the table of a planer be considered. The cross-section 
is almost universally of the form shown in Fig. 155. This is evi- 
dently a form that would yield easily to a force tending to twist it, 
or to a force acting in a vertical plane tending to bend it. Such 
forces may be brought upon it by " strapping down work," or by the 

* See Richards' " Manual of Machine Construction." 



144 MACHINE DESIGN. 

support of heavy pieces upon centres. Thus in Fig. 156 the heavy 
piece E is supported between the centres. For proper support the 
centres need to be screwed in with a considerable force. This 
causes a reaction tending to separate the centres and, to bend the 
table between C and D. ,As a result of this, the Vs on the table no 
longer have a bearing throughout the entire surface of the guides on 
the bed, but only, touch near the ends, the pressure is concentrated 
upon small surfaces, the lubricant is squeezed out, the Vs and 
guides are "cut," and the planer is rendered incapable of doing ac- 
curate work. If the table were made of the box form shown in 
Fig. 157, with partitions at intervals throughout its length, it 
would be far more capable of maintaining its accuracy of form un- 
der all kinds of stress, and would be more satisfactory for the pur- 
pose for which it is designed.* 

The bed of a planer is usually of the form shown in section in 
Fig. 158, the side members being connected by " cross girts" at in- 
tervals. This is evidently not the best form to resist flexure and 
torsion, and a planer bed may sustain both, either by reason of 
improper support, or because of changes in the form of foundation. 
If the bed were of box section with cross partitions, it would sus- 
tain greater stress without undue yielding. Holes could be left in 
the top and bottom to admit of supporting the core in the mould ; 
to serve for the removal of the core sand ; and to render accessible 
the gearing and other mechanism inside of the bed. 

This same reasoning applies to lathe beds. They are strained 
transversely by force tending to separate the centres, as in the case 
of " chucking " ; torsionally by the reaction of a tool cutting the 
surface of a piece of large diameter ; and both torsion and flexure 
may result, as in the case of the planer bed, from an improperly 
designed or yielding foundation. The box form would be the best 
possible form for a lathe bed ; some difficulties in adaptation, how- 
ever, have prevented its extended use as yet. 

These examples illustrate principles that are of very broad ap- 
plication in the designing of machines. 

* Prof. Sweet has designed and constructed such a table for a large mill- 
ing machine. 



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FORM OF MACHINE PARTS AS DICTATED BY STRESS. 145 

128. Often in machines there is a part that projects either verti- 
cally or horizontally and sustains a transverse stress ; it is a canti- 
lever, in fact. If only transverse stress is sustained, and the thick- 
ness is uniform, the outline for economy of material is parabolic. 
In such a case, however, the outline curve of the member should 
start from the point of application of the force, and not from the 
extreme end of the member, as in the latter case there would be an 
excess of material. Thus in A, Fig. 159, P is the extreme position 
at which the force can be applied. The parabolic curve a is drawn 
from the point of application of P. The end of the member is sup- 
ported by the auxiliary curve c. The curve b drawn from the end 
gives an excess of material. The curves a and c may be replaced 
by a single continuous curve as in 0, or a tangent may be drawn 
to a at its middle point as in B, and this straight line used for the 
outline ; the excess of material being slight in both cases. Most 
of the machine members of this kind, however, are subjected also 
to other stresses. Thus the " housings " of planers have to resist 
torsion and side flexure. They are very often supported by two 
members of parabolic outline ; and, to insure the resistance of the 
torsion and side flexure, these two members are connected at their 
parabolic edges by a web of metal that really converts it into a box 
form. Machine members of this kind may also be supported by a 
brace, as in D. The brace is a compression member and may be 
stiffened against buckling by a " web" as shown, or by an auxiliary 
brace. 

J 9 



CHAPTER XV. 



MACHINE SUPPORTS. 



129. The Single Box Pillar Support is best and simplest for ma* 
chines whose size and form admit of its use. When a support is a 
single continuous member, its design should be governed by the 
following principles: 

I. The amount of material in the cross-section is determined by 
the intensity of the load. If vibrations are also to be sustained, 
the amount of material must be increased for this purpose. 

II. The vertical centre line of the support should coincide with 
the vertical line through the centre of gravity of the part supported. 

III. The vertical outlines of the support should taper slightly 
and uniformly on all sides. If they were parallel they would 
appear nearer together at the bottom. 

IV. The external dimensions of the support must be such that 
the machine has the appearance of being in stable equilibrium. 
The outline of all heavy members of the machine supported must 
be either carried without break to the foundation, or if they over- 
hang, must be joined to the support by means of parabolic outlines, 
or by the straight lines of the brace form. 

Illustration. — In Fig. 160 the first three principles may be ful- 
filled, but there is an appearance of instability. It is because the 
outline of the "housing" overhangs. It should be carried to the 
foundation without break in the continuity of the metal, as in 
Fig. 161. 

130. When the support is divided up into several parts, modi- 
fication of these principles becomes necessary, as the divisions 
require separate treatment. This question may be illustrated by 



MACHINE SUPPORTS. 147 

lathe supports. In Fig. 162 are shown three forms of support for a 
lathe, seen from the end. For stability the base needs to be broader 
than the bed. In A the width of base necessary is determined and 
the outlines are straight lines. The unnecessary material is cut 
away on the inside, leaving legs, which are compression members of 
correct form. The cross brace is left to check any tendency to 
buckle. For convenience to the workmen it is desirable to narrow 
this support somewhat without narrowing the base. The cross 
brace converts the single compressson member into two compression 
members. It is allowable to give these different angles with the 
vertical. This is done in B and the straight lines are blended into 
each other by a curve. C shows a common incorrect form of lathe 
support, the compression members from the cross brace downward 
being curved. There is no reason for this curved form. It is less 
capable of bearing its compressive load than if it were straight, and 
is no more stable than the form b, the width of base being the same. 

Consider the lathe supports from the front. Four forms are shown 
in Fig. 163. If there were any force tending to move the bed of the 
lathe endwise the forms B and C would be allowable. But there is 
no force of this kind, and the correct form is the one shown in D. 
Carrying the foot out as in A, B, and C, increases the distance 
between supports (the bed being a beam with end supports and the 
load between); this increases the deflection and the fibre stress due 
to the load. This increase in stress is probably not of any serious 
importance, but the principle should be regarded or the appearance 
of the machine will not be right. If the supports were joined by a 
cross member, as in Fig. 164, they would be virtually converted into 
a single support, and should then taper from all sides. 

131. If a machine be supported on a single box pillar, change in 
the form of the foundation cannot induce stress in the machine 
frame tending to change its form. If, however, the machine is sup- 
ported on four or more legs the foundation might sink away from 
one or more of them and leave a part unsupported. This might 
cause torsional or flexure stress in some part of the machine, which 
might change its form, and interfere with the accuracy of its action. 



148 MACHINE DESIGN. 

But if the machine be . supported on three points this cannot occui; 
because, if the foundation should sink under any one of the sup- 
ports, the support would follow and the machine would still rest on 
three points. When it is possible, therefore, a machine which can- 
not be carried on a single pillar should be supported on three 
points. Many machines are too large for three-point support, and 
the resource is to make the bed, or part supported, of box section 
and so rigid that even if some of the legs should be left without 
foundation, the part supported would still maintain its form. More 
supports are often used than are necessary. Thus, if a lathe has 
two pairs of legs like those shown in B, Fig. 162, and these are bolted 
firmly to the bed, there will be four points of support. But if, as 
suggested by Professor Sweet, one of these pairs be connected to the 
bed by a pin so that the support and the bed are free to move, 
relatively to each other, about the pin, as in Fig. 165, then this is 
equivalent to a single support, and the bed will have three points of 
support, and will maintain its form independently of any change 
in the foundation. This is of special importance when the ma- 
chines are to be placed upon yielding floors. 

132. Fig. 166 shows another case in which the number of sup- 
ports may be reduced without sacrifice. In A three pairs of legs 
are used. There are therefore six points of support. In B two 
pairs of legs are used and one may be connected by a pin, and there 
will be but three points of support. The chance of the bed being 
strained from changing foundation, has been reduced from 6 in A 
to in B. The total length of bed is 12 ft., and the unsupported 
length is 6 ft. in both cases. 

133. Figs. 167 and 168 show correct methods of support for small 
lathes and planers, due to Professor Sweet. In Fig. 167 the lathe 
" head stock" has its outlines carried to the foundation by the box 
pillar ; a represents a pair of legs connected to the bed by a pin 
connection, and instead of being placed at the end of the bed it is 
moved in somewhat, the end of the bed being carried down to the 
support by a parabolic outline. The unsupported length of bed is 
thereby decreased, the stress on the bed is less, and the bed will 




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MACHINE SUPPORTS. 149 

maintain its form regardless of any yielding of the floor or founda- 
tion. In Fig. 168 the housings, instead of resting on the bed as is 
usual in small planers, are carried to the foundation, forming two 
of the supports ; the other is at a and has a pin connection with the 
bed, which being thus supported on three points cannot be twisted 
or flexed by a yielding foundation. 



CHAPTER XVI. 



MACHINE FKAMES 



134. Fig. 169 shows an open side frame, such as is used for punch- 
ing and shearing machines. During the action of the punch or shear 
a force is applied to the frame tending to separate the jaws. This 
force may be represented in magnitude, direction, and line of action 
by P. It is required to find the resulting stresses in the three sec- 
tions AB, CD, and EF. Consider AB. Let the portion above this 
section be taken as a free body. The force P, Fig. 170, and the op- 
posing resistances to deformation of the material at the section AB, 
are in equilibrium. Let H be the projection of the gravity axis of 
the section AB, perpendicular to the paper. Two equal and oppo- 
site forces, P x and P 2 , may be applied at H without disturbing the 
equilibrium. Let P x and P 2 be each equal to P, and let their line of 
action be parallel to that of P. The free body is now subjected to 
the action of an external couple, PI, and an external force, P x . The 
couple produces flexure about H, and the force P x produces tensile 
stress in the section AB. The flexure results in a tensile stress 
varying from a maximum value in the outer fibre at A to zero at H, 
and a compressive stress varying from a maximum in the outer 
fiber at B to zero at H. This may be shown graphically at JK. 
The ordinates of the line LM represent the varying stress due to 
flexure ; while ordinates between LM and NO represent the uni- 
form tensile stress. This latter diminishes the compressive stress 
at B, and increases the tensile stress at A. The tensile stress per 
square inch at A therefore equals S -j- S l ; where S equals the unit 
fibre stress due to flexure at A, and S x equals the unit tensile stress 

Pic P 

due to P v Now S — -j- } and S x = -. ; in which c = the distance 



MACHINE FRAMES. 151 

from the gravity axis to the outer fibre = AH, and J = the mo- 
ment of inertia of the section about H, and A = area of the cross- 
section AB. 

Let it be required to design the frame of a machine to punch f" 
holes in \" steel plates, 18" from the edge. The surface resisting 
the shearing action of the punch = n X j" X V = 1'17 sq. in. 
The ultimate shearing strength of the material is say 50000 pounds 
per square inch. The total force, P, which must be resisted by the 
punch frame = 50000 X 117 =58500 pounds. 

135. The material and form for the frame must first be selected. 
The form is such that forged material is excluded, and difficulties of 
casting and high cost exclude steel casting. The material, therefore, 
must be cast iron. Often the same pattern is used both for the 
frame of a punch and shear. In the latter case when the shear 
blade begins and ends its cut the force is not applied in the middle 
plane of the frame, but considerably to one side, and a torsional 
stress results in the frame. Combined torsion and flexure are best 
resisted by members of box form. The frame will therefore be 
made of cast iron and of box section. The dimension AB may be 
assumed so that its proportion to the " reach" of the punch appears 
right : the width and thickness of the cross-section may also be 
assumed. From these data the maximum stress in the outer fibre 
may be determined. If this is a safe value for the material used 
the design will be right. 

136. Let the assumed dimensions be as shown in Fig. 171. Then 

A = b x d x — b 2 d 2 = 78 sq. in. 

T bjd' — b.,d./' 

12~~~ 

= 3000 bi-quadratic inches, nearly. 

c = d t 4-2 == 9"; I = the reach of the punch -j- c = 27"; P = 
58500 lbs., as determined above. Then 



152 MACHINE DESIGN. 

P 58500 

„ Pic 58500 X 27 X 9 . oen 
S = T = 3000~ =486 °- 

S l J r S = 5630 = maximum stress in the section. 

The average strength of cast iron such as is used for machinery 
castings, is about 20000 lbs. per square inch. The factor of safety 
in the case assumed equals 20000 -r- 5630 = 3'5. This is too small. 
There are two reasons why a large factor of safety should be used 
in this design : I. When the punch goes through the plate the yield- 
ing is sudden and a severe stress results. This stress has to be sus- 
tained by the frame, which for other reasons is made of unresilient 
material. II. Since the frame is of cast iron, there will necessarily 
be shrinkage stresses which the frame must sustain in addition to 
the stress due to external forces. These shrinkage stresses cannot 
be calculated and therefore can only be provided against by a large 
factor of safety. 

Cast iron is strong to resist compression and weak to resist ten- 
sion, and the maximum fibre stress is tension on the inner side. 
The metal can therefore be more satisfactorily distributed than in 
the assumed section, by being thickened where it sustains tension, 
as at a, Fig. 172. If, however, there is a very thick body of metal 
at a, sponginess and excessive shrinkage would result. The form 
B would be better, the metal being arranged for proper cooling and 
for the resisting of flexure stress. 

137. Dimensions may be assigned to a section like B and the 
cross-section may be checked for strength as before. See Fig. 173. 
GrGr, a line through the centre of gravity of the section, is found to 
be at a distance of 7" from the tension side. The required values 
are as follows : c = 7"; I = reach of punch + c = 18 + 7 = 25"; 
A = 156*5 sq. in.; 7 = 5000 bi-quadratic inches, nearly; P — 58500 
lbs. 




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MACHINE FRAMES. 158 

rvu a P 58500 on A 11 

Ihen ^ ________ 374 lbs. 

Pic 58500X25X7 oft .» .. 

S = T= -M)0 = 20471bs - 

iST- —J— jSPj= 2421 lbs. = maximum fibre stress in the section. The 
factor of safety = 20000 -J- 2421 = 8-25. This section, therefore, 
fulfills the requirement for strength, and the material is well 
arranged for cooling with little shrinkage and without spongy 
spots. The gravity axis may be located, and the value of / deter- 
mined by graphic methods. See Hoskins' "Graphic Statics." 

138. Let the section CD, Fig. 169, be considered. Fig. 174 shows 
the part at the left of CD free. K is the projection of the gravity 
axis of the section. As before, put in two opposite forces 1\ and P 4 , 
equal to each other and to P, and having their common line of 
action parallel to that of P, at a distance l x from it. P and P 4 now 
form a couple, whose moment — Pl x , tending to produce flexure 
about K. P 3 must be resolved into two components, one P s J\ at 
right angles to the section considered, tending to produce tensile 
stress; and the other JK, parallel to the section, tending to produce 
shearing stress. The greatest unit tensile stress in this section will 
equal the sum of that due to flexure and that due to tension == 

S + Sr = Y + P f- 

IK 

The greatest unit shear — S s — — p 

139. In the section FE, Fig. 169, which is parallel to the line of 
action of P, equal and opposite forces, each = P, may be introduced, 
as P 6 and P 6 . P and P 6 will then form a couple* with an arm l„ 
and P 5 will be wholly applied to produce shearing stress. The 
maximum unit tensile stress in this section will be that due to 
flexure, S Pic, : /, and the maximum unit shear will be S 

P -f- A. Any section may be thus checked. 



154 MACHINE DESIGN. 

140. The dimensions of several sections being found, the outline 
curve bounding them should be drawn carefully, to give good ap- 
pearance. The necessary modifications of the frame to provide for 
support, and for the constrainment of the actuating mechanism, 
may be worked out as in Fig. 175. A is the pinion on the pulley 
shaft from which the power is received ; B is the gear on the main 
shaft; C, D, and G are parts of the frame added to supply bear- 
ings for the shafts ; E furnishes the guiding surfaces for the punch 
" slide." The method of supporting the frame is shown, the support 
being cut under at F for convenience to the workman. The parts 
C, D, E, and G can only be located after the mechanism train has 
been designed. 

141. Slotting Machine Frame. — See Fig. 176. It is specified that 
the slotter shall cut at a certain distance from the edge of any 
piece, and the dimension AH is thus determined. The table G 
must be held at a convenient height above the floor, and RK must 
provide for the required range of " feed." K is cut under for con- 
venience to the workman, and carried to the floor line as shown. 
It is required to " slot " a piece of given vertical dimension, and the 
distance from the surface of the table to E is thus determined. Let 
the dimension LM be assumed so that it shall be in proper propor- 
tion to the necessary length and height of the machine. The curves 
LS and MT may be drawn for bounding lines of a box frame to 
support the mechanism. M should be carried to the floor line as 
shown, and not cut under. None of the part DNE, nor that which 
serves to support the cone and gears on the other side of the frame, 
should be made flush with the surface LSTM, because nothing 
should interfere with the continuity of the curves LS and TM. The 
supporting frame of a machine should be clearly outlined, and other 
parts should appear as attachments. The member VW should be 
designed so that its inner outline is nearly parallel to the outline of 
the cone pulley, and should be joined to the main frame by a curve. 
The outer outline should be such that the width of the member 
increases slightly from W to V, and should also be joined to the 
main frame by a curved outline. In any cross-section of the frame, 



MACHINE FRAMES. 155 

as XX, the amount of metal and its arrangement may be controlled 
by the core. It is dictated by the maximum force, P, which the 
tool can be required to sustain. The tool is carried by the slider of 
a slider crank chain. Its velocity varies, therefore, from a max- 
imum near mid-stroke, to zero at the upper and lower ends of its 
stroke. The belt which actuates the mechanism runs on one of the 
steps of the cone pulley, at a constant velocity. Suppose that the 
tool is set (accidentally) so that it strikes the table just before the 
slider has reached the lower end of its stroke. The resistance, R, 
offered by the tool to being stopped, multiplied by its (very small) 
velocity, equals the difference of belt tension multiplied by the belt 
velocity (friction and inertia neglected). R, therefore, would vary 
inversely as the slider velocity, and hence may be very great. Its 
maximum value is indeterminate. A "breaking piece " may be 
put in between the tool and the crank. Then when R reaches a 
certain value, the breaking piece fails. The stress in the stress- 
members of the machine is thereby limited to a certain definite 
value. From this value the frame may be designed. Let P = up- 
ward force against the tool when the breaking piece fails. Let 
I — the horizontal distance from the line of action P to the gravity 
axis of the section XX. Then the section XX sustains flexure stress 
caused by the moment PI, and tensile stress equal to P. The max- 
imum unit stress in the section = 

A section may be assumed and checked for safety, as for the punch. 
142. Stresses in the Frame of a Side-Crank Steam Engine. — Fig. 
177 is a sketch in plan of a side-crank engine of the "girder bed" 
type. The supports are under the cylinder C, the main bearing E, 
and the out-board bearing D. A force P is applied in the centre line 
of the cylinder, and acts alternately toward the right and toward 
the left. In the first case it tends to separate the cylinder and 
main shaft ; and in the second case it tends to bring them nearer 



156 MACHINE DESIGN. 

together. The frame resists these tendencies with resulting inter- 
nal stresses. 

Let the stresses in the section AB be considered. The end of 
the frame is shown enlarged in Fig. 178. If the pressure from the 
piston is toward the right, the stresses in AB will be : I. Flexure 
due to the moment PI, resulting in tensile stress below the gravity 
axis N, with a maximum value at b, and a compressive stress above 
N with a maximum value at a. II. A direct tensile stress, — P, 
distributed over the entire section, resulting in a unit stress = 
P -f- A ■== S lbs. per sq. in. This is shown graphically at n, Fig. 178. 
a l b l is a datum line whose length equals AB. Tensions are laid off 
toward the right and compressions toward the left. The stress due 
to flexure varies directly as the distance from the neutral axis iV 13 
being zero at i\ 7 r If, therefore, b 1 c 1 represents the tensile stress in 
the outer fibre, then c 1 k 1 drawn through N will be the locus of the 
ends of horizontal lines, drawn through all points of a t b v represent- 
ing the intensity of stress, in all parts of the section, due to flexure. 
If c x d x represent the unit stress due to direct tension, then, since this 
is the same in all parts of the section, it will be represented by the 
horizontal distance between the parallel lines c l k l and d l e 1 . This 
uniform tension increases the tension b 1 c l due to flexure, causing 
it to become b^ ; and reduces the compression Jc i a v causing it 
to become e l a 1 . The maximum stress in the section is therefore 
tensile stress in the lower outer fibre, and is equal to b 1 d 1 . 

When the force P is reversed, acting toward the left, the stresses 
in the section are as shown at m : compression due to flexure in the 
lower outer fibre equal to c 2 b 2 ; tension due to flexure in the upper 
outer fibre equal to a 2 k 2 ; and uniform compression over the en- 
tire surface equal to d 2 c r This latter increases the compression 
in the lower outer fibre from b 2 c 2 to b 2 d 2 , and decreases the tension 
in the upper outer fibre from a 2 k 2 to a 2 e 2 . The maximum stress in 
the section is therefore compression in the lower outer fibre equal to 
b 2 d. r The maximum stress, therefore, is always in the side of tire 
frame next to the connecting-rod. 

If the gravity axis of the cross-section be moved toward the 




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MACHINE FRAMES. 157 

connecting-rod, the stress in the upper outer fibre will be increased 1 ; 
and that in the lower outer fibre will be proportionately, decreateed; 
The gravity axis may be moved toward the connecting-rod by- 
increasing the amount of material in the lower part of the cross- 
section and decreasing it in the upper parti 

The stresses in any other section nearer the cylinder will be due 
to the same force, P, as before ; but the moment tending to produce 
flexure will be less, because the lever arm of the moment is less and 
the force constant. 

143. Suppose the engine frame to be of the type which is con- 1 
tinuous with the supporting part as shown in Fig. 179. Let Pig. 
180 be a cross-section, say at AB. is the centre of the cylinder. 
The force P is applied at this point perpendicular to the paper. C 
is the centre of gravity of the section (the intersection of two 
gravity axes perpendicular to each other, found graphically). Join' 
C and 0, and through C draw XX perpendicular to CO. Then XX 
is the gravity axis about which flexure will occur.* The dangerous 
stress will be at F, and the value of c will be the perpendicular dis- 
tance from F to XX. The moment of inertia of the cross-section 
about XX may be found, =-- / ; I, the lever arm of P, = 0C-. The 
stress at F, S -f S x , must be safe value. 

S = ~j in known terms. 

p 

S, = — = r-, in known terms. 

area of sec n 

*This is not strictly true. If OC is a diameter of the " ellipse of inertia," 
flexure will occur about its conjugate diameter. If the section of the engine 
frame is symmetrical with respect to a vertical axis, OC is vertical, and its 
conjugate diameter XX is horizontal. Flexure would occur about XX, and 
the angle between OC and XX would equal 90°. As the section departs from 
symmetry about a vertical, XX, at right angles to OC, departs from OC's con- 
jugate, and hence does not represent the axis about which flexure occurs. 
In sections like Fig. 178, the error from making ft = 90° is unimportant. 
When the departure from symmetry is very great, however, OC's conjugate 
should be located and used as the axis about which flexure occurs. For 
method of drawing " ellipse of inertia " see Hoskins' " Graphic Statics." 



158 MACHINE DESIGN. 

144. Closed Frames. — Fig. 181 shows a closed frame. The mem- 
bers G and H are bolted rigidly to a cylinder C at the top, and to a 
bed plate, DD, at the bottom. A force P may act in the centre line, 
either to separate D and 0, or to bring them nearer together. The 
problem is to. design G, H, and D for strength. If the three mem- 
bers were " pin connected," see Fig. 182, the reactions of C upon 
A and B at the pins would act in the lines EF and GH. Then if P 
acts to bring and C nearer together, compression results in A, the 
line of action being EF; compression results in B, the line of action 
being GH. These compressions being in equilibrium with the force 
P, their magnitude may be found by the triangle of forces. From 
these values A and B may be designed. C is equivalent to a beam 
whose length is I, supported at both ends, sustaining a transverse 
load P, and tension equal to the horizontal component of the com- 
pression in A or B. The data for its design would therefore be 
available. Reversing the direction of P reverses the stresses ; the 
compression in A and B becomes tension ; the flexure moment 
tends to bend C convex downward instead of upward, and the ten- 
sion in C becomes compression. 

145. But when the members are bolted rigidly together, as in 
Fig. 181, the lines of the reactions are indeterminate. Assump- 
tions must therefore be made. Suppose that G is attached to D by 
bolts at E and A. Suppose the bolts to have worked slightly loose, 
and that P tends to bring C and D nearer together. There would 
be a tendency, if the frame yields at all, to relieve pressure at E 
and to concentrate it at A. The line of the reaction would pass 
through A and might be assumed to be perpendicular to the surface 
AE. Suppose that P is reversed and that the bolts at A are 
loosened, while those at E are tight. The line of the reaction 
would pass through E, and might be assumed to be perpendicular 
to EA. MN is therefore the assumed line of the reaction, and the 
intensity = P -h 2. In any section of G, as XX, let K be the pro- 
jection of the gravity axis. Introduce at K, two equal and opposite 
forces, equal to R and with their lines of action parallel to that of 
R. Then in the section there is flexure stress due to the flexure 



to. 



n 



/To. /a/ . 




K l 



T~rv 



MACHINE FRAMES. 159 

moment Rl, and tensile stress due to the component of R 2 perpen- 
dicular to the section, = R 3 . Then the maximum stress in the sec^ 

tion =S+ S v 

R. A Rlc 

A section may be assumed, and A, I, and c become known ; 
the maximum stress also becomes known, and may be compared 
with the ultimate strength of the material used. 

Obviously this resulting maximum stress is greater when the 
line of the reaction is MN, than when it is KL. Also it is greater 
when MN is perpendicular to EA, than if it were inclined more 
toward the centre line of the frame. The assumptions therefore 
give safety. If the force P could only act downward, as in a steam 
hammer, KL would be used as the line of the reaction. 

146. The part D in the bolted frame, is not equivalent to a beam 

with end supports and a central load like 0, Fig. 182, but more 

nearly a beam built in at the ends with central load ; and it may be 

so considered, letting the length of the beam equal the horizontal 

distance from E to F, — I. Then the stress in the mid-section will 

PI Pic 

be due to the flexure moment -^-,and the maximum stress = S=—j. 

o o 1 

The values c and / may be found for an assumed section, and S 
becomes known. 



INDEX 



Addendum, 148. 

Angularity of connecting-rod, 20. 

Annular gears, 50. 

Belts, 75. 

centrifugal force of, 89. 

design of, 81. 

transmission by, 75. 
Bevel gears, 59. 
Bolts and screws, 130. 

and screws, design of, 131. 

cross-section of, to resist tension, 
135. 

elongation of, 134. 

fastenings to hold steam chest 
cover, 132. 

tendency to loosen nuts, 136. 

Cams, 72. 

Cantilever in machines, 145. 
Centro of two gears, 40. 
Centros of relative motion, 13. 

in compound mechanism, 14. 
Complete constrainmentof motion, 5. 
Cone pulleys, 78. 
Conservation of energy, 1. 
( lonstrained motion, 3. 
Cotter, 138. 
Cycloidal curves, 42. 

teeth, 60. 



Energy in machines, 28. 

is transferred in time, 29. 

Feathers, or splines, 137. 

rules for, 137. 
Fly-wheel, pump, 97. 

steam engine, 99. 

design of, 93. 
Force problems, 29. 

in the steam engine, 35 
Form of machine parts, 140. 

in stress, tension, or compres- 
sion, 140. 

in transverse stress, 141. 

in torsional stress, 142. 
Frame of machine, 143. 

slotting machine, 154. 

closed, 158. 
Free motion, 3. 
Function of machines, 2. 

Gears : angular velocity ratio, 70. 
annular, 50. 
backlash, 48. 
bevel, 59. 

bevel, design of, 63. 
clearance, 48. 

compound spur gear chains, 69. 
definitions, 45, 47. 
design of worm, 67. 



162 



INDEX. 



Gears, diametral pitch, 48. 
face, 48. 
formulas, 55. 
interchangeable, 51. 
interchangeable involute, 53. 
laying out, 54. 
skew bevel, 64. 
solution from other data, 68. 
strength of teeth, 56. 
total depth, 48. 
tooth, design of, 57. 
spiral, 64. 
working depth, 48. 
worm, 65. 

Generating circle, 43. 
Guides, 128. 

Higher pairs, 39. 

Independent constrainment of mo- 
tion, 13. 

Instantaneous motion and instanta- 
neous centres or centros, 7. 

Interchangeable gears, 43. 

Involute teeth, 61. 
tooth outlines, 46. 

Journals, design of , 111. 

allowable pressure, 112. 
bearings and boxes, 120. 
direction of motion, 112. 
frictional resistance of, 113. 
lubrication of, 123. 
oiling, 124. 

pressure in thrust, 118. 
radiation of, 114. 
stationary, 124. 
thrust, 117. 

Keys, as a means for preventing rela- 
tive rotation, 137. 



Keys, rules for, 138. 

Kinds of motion in machines, 6. 

Lever crank chain, 13. 
Linkages or motion chains ; mechan- 
isms, 11. 
Location of centros, 12. 
Loci of centros or centroids, 9. 

Machine frames, 150. 
Machine supports, 147. 
Machinery of application, 2. 
Machinery of transmission, 2. 
Means for preventing relative rota- 
tion, 137. 
Motion independent of force, 6. 

Non-circular wheels, 59. 

Open frame design, 152. 

side frames, 150. 
Outline of machine frame, 154. 

Pairs of motion elements, 10. 

Parallel or straight line motions, 37. 

Passive resistance, 3. 

Pitch point, 41. 

Prime mover, 2. 

Pulleys, cone, graphical method, 79. 

Racks, 48. 

Rate of doing work, 28. 

Ratio of a quick return, 21. 

Reduction of the number of sup- 
ports, 148. 

Relative linear velocity in same 
link, 17. 
linear velocity not in same 

link, 18. 
motion, 6. 

Rigid body, 7. 

Riveted joints, 100. 



INDEX. 



163 



Riveted joints, butt, 101. 
lap, 101. 

margin of rivets, 106. 
pitch of rivets, 105. 
table for, 103. 

table for rivet diameters, 104. 
table of efficiency, 107. 

Shrink and force fits, 139. 
Slider crank chain, 12. 

mechanism, 12. 

and guide of unequal length, 127 
Sliding surfaces, 126. 
Slotted cross-head mechanism, 14. 
Slotting machine frame, 154. 
Solution of a quick return, 23. 
Steam engine, stresses in, 155. 



Stresses in the frame of a steam en- 
gine, 155. 
Support divided into several parts, 
146. 
for lathes, 148. 
single box pillar, 147. 

Table for use in designing belts, 85. 

Tooth outlines, 41 

Toothed wheels, or gears, 39. 

Vector, 16. 
Velocity, 15. 

of cutting tools, 21. 

Whitworth quick return mechanism, 
25. 



021 225 325 A 




